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show that the function f(x)= sqrt (x^2 +1) satisfies the 2 hypotheses of the Mean Value Theorem

show that the function f(x)= sqrt (x^2 +1) satisfies the 2 hypotheses of the Mean Value Theorem on the interval [0, sqrt 8]. then, find any values of c guarenteed by the theorem.

Research, Knowledge and Information :


The Mean Value Theorem - Department of Mathematics


ex. Show that the function f (x) ... of c that satisfies the M.V.T. on the interval [−1, 1]. ... (x) ≤ 2 for all x, by the Mean Value Theorem the average
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How do you verify that the function f(x)= (sqrt x)- 1/3 x ...


... (sqrt x)- 1/3 x# satisfies the three hypothesis of Rolles's ... Derivative Mean Value Theorem for Continuous Functions. ... f satisfies the hypotheses ...
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Use the Mean Value Theorem to show that sqrt(1 + x) < 1 + (1 ...


Use the Mean Value Theorem to show that sqrt(1 + x ... Apply the Mean Value Theorem to the function f(x ... Verify that the function f satisfies the hypotheses of ...
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Show That The Function F(x) = Sqrt(x) Satisfies Th... | Chegg.com


Answer to Show that the function f(x) = sqrt(x) satisfies the hypotheses of the mean value theorem on the interval [0, 4]. Find al...
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The Mean-Value Theorem


The Mean Value Theorem is one of the most important theoretical tools in ... let f(x) be a differentiable function on an interval I, with f ... Exercise 2. Show that
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Mean Value Theorem Help! (295063) | Wyzant Resources


Verify the function f(x)=1/x satisfies the hypothesis of the mean value theorem on [1,3]. Then find all numbers c that satisfy the conclusion of the mean value theorem.
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Mean-Value Theorem - UMass Dartmouth


The Mean Value Theorem Let f be a function that satisfies the following ... = 2x-x^2 which satisfies both of the hypotheses of the Mean Value Theorem on the ...
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Worksheet 21: The Mean Value Theorem - UCB Mathematics


Worksheet 21: The Mean Value Theorem Russell Buehler [email protected] 1. Verify that f(x) = x3 x2 6x+ 2 satis es the hypotheses www.xkcd.com of Rolle’s theorem for ...
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Suggested Questions And Answer :


show that the function f(x)= sqrt (x^2 +1) satisfies the 2 hypotheses of the Mean Value Theorem

f(x) = sqrt(x^2 + 1) ; [(0, sqrt(8)] Okay, so for the Mean Value Theorem, two things have to be true: f(x) has to be continuous on the interval [0, sqrt(8)] and f(x) has to be differentiable on the interval (0, sqrt(8)). First find where sqrt(x^2 + 1) is continous on. We know that for square roots, the number has to be greater than or equal to zero (definitely no negative numbers). So set the inside greater than or equal to zero and solve for x. You'll get an imaginary number because when you move 1 to the other side, it'll be negative. So, this means that the number inside the square root will always be positive, which makes sense because the x is squared and you're adding 1 to it, not subtracting. There would be no way to get a negative number under the square root in this situation. Therefore, since f(x) is continuos everywhere, (-infinity, infinity), then f(x) is continuous on [0, sqrt(8)]. Now you have to check if it is differentiable on that interval. To check this, you basically do the same but with the derivative of the function. f'(x) = (1/2)(x^2+1)^(-1/2)x2x which equals to f'(x) = x/sqrt(x^2+1). So for the derivative of f, you have a square root on the bottom, but notice that the denominator is exactly the same as the original function. Since we can't have the denominator equal to zero, we set the denominator equal to zero and solve to find the value of x that will make it equal to zero. However, just like in the first one, it will never reach zero because of the x^2 and +1. Now you know that f'(x) is continous everywhere so f(x) is differentiable everywhere. Therefore, since f(x) is differentiable everywhere (-infinity, infinity), then it is differentiable on (0, sqrt(8)). So the function satisfies the two hypotheses of the Mean Value Theorem. You definitely wouldn't have to write this long for a test or homework; its probably one or two lines of explanation at most. But I hope this is understandable enough to apply to other similar questions!
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Rolles theorem on f(x) = Xsqrt(64-X^2) on the interval [-8,8]?

your derivative should have been sqrt(64-x^2) - x^2/sqrt(64-x^2) The minus sign coming from the derivative of (-x^2) Setting the derivative to zero, sqrt(64-x^2) - x^2/sqrt(64-x^2) = 0   multiply both terms by sqrt(64 - x^2) (64 - x^2) - x^2 = 0 64 = 2x^2 32 = x^2 x = +/- 4.sqrt(2) Answer: option b
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How to Find Square Root

98=49*2, so sqrt(98)=sqrt(49)*sqrt(2)=7sqrt(2)=7*1.4142=9.8994 approx. There's another way using the binomial theorem. 98=100-2=100(1-0.02). sqrt(100)=10 so sqrt(98)=10(1-0.02)^(1/2) because square root is the same as power 1/2. (1+x)^n expands to 1+nx+(n(n-1)/1*2)x^2+(n(n-1)(n-2)/1*2*3)x^3+... Putting n=1/2 and  x=-0.02, we get sqrt(98)=10(1-0.02)^(1/2)=10[1-(1/2)0.02+((1/2)(-1/2)/2)0.0004+...]. This gives us: 10(1-0.01-0.00005+...)=10*0.98995=9.8995. A third method is to use an iterative process, which means you keep repeating the same action over and over again. Look at this: x=10-(2/(10+x)). If we solve for x we get x=sqrt(98); but we're going to find x in an iterative way. Start with x=0 and work out the right hand side: 10-2/10=9.8. This gives us a new value for x, 9.8, which we feed back into the right hand side: 10-(2/(10+9.8))=10-2/19.8=9.8989..., giving us another value for x, 9.8989... which we feed back into the right hand side: 10-(2/(10+9.8989...))=9.89949..., giving us yet another value for x and so on. Very quickly we build up accuracy with each x. You can do this on a calculator, a basic one that doesn't even have square roots, using the memory to hold values for you. Here's a very simple program, where STO means store in memory (if your calculator doesn't have STO use MC (memory clear) followed by M+ (add to memory)); MR means read memory (the steps show what calculator keys to press in order; / may be ÷ on your calculator): 0= +10=STO 10-2/MR= GO TO STEP 2 OR STOP (display should show the answer for sqrt(98)) Note: In STEP 3 the division must be carried out before subtracting from 10, otherwise you get the wrong answer. If your calculator doesn't do this you need to replace STEP 3 with: 0-2=/MR=+10= You should only have to go round the loop a few times before you get a really accurate result. To find the square root of 2 directly the iteration equation is x=1+1/(1+x) and the program is: 0= +1=STO 1/MR+1= GO TO STEP 2 OR STOP STEP 3 should work on all calculators.
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Show that the plane 3x+.... (Analytical Geometry)

What I'm trying to do here is to establish the gradients of the ellipsoid and the plane. The plane has only one gradient (in any particular direction) whereas the ellipsoid gradient varies over the surface. Where the ellipsoid surface has the same gradient as the fixed gradient of the plane gives us point(s) on the surface. This means that the plane and the gradient (tangent) of the ellipsoid are parallel, when the gradient of the ellipsoid is measured in the same direction as the plane. The plane touches the ellipsoid only if they meet, that is, the coordinates of the ellipsoid at the surface match coordinates lying on the plane.  [In the following, I may not be able to give you the whole proof, due to my unfamiliarity with this type of problem, but I have an intuition regarding the essence of the solution. I hope that you will be able to glean enough for you to pursue the solution yourself.] Let w=-sqrt(7)(1-2x^2-y^2)+2z^2. grad(w)=<4xsqrt(7),2ysqrt(7),4z>. This is derived from the differentials of the variable terms. grad(w) at point P(a,b,c) is grad(w)|P=<4asqrt(7),2bsqrt(7),4c>, making the tangent plane: 4asqrt(7)(x-a)+2bsqrt(7)(y-b)+4c(z-c)=0 or 4asqrt(7)x+2bsqrt(7)y+4cz=4a^2sqrt(7)+2b^2sqrt(7)+4c^2. The point P also lies on the plane: 3a+b-c=4, if the ellipsoid touches the plane. So, c=3a+b-4. P must lie on the ellipsoid, so if a and b are set, c^2=z^2=(1-2a^2-b^2)sqrt(7)/2. [However, if z=0, it is easier to find values for b and c on the surface of the ellipsoid. Such values are given by 2x^2+y^2=1, an ellipse itself. So P(a,sqrt(1-2a^2),0) is such a point. The tangential plane is 4asqrt(7)x+2sqrt(7(1-2a^2))y=4a^2sqrt(7)+2(1-2a^2) and 3a+sqrt(1-2a^2)-4=0 if P lies on the plane. So sqrt(1-2a^2)=4-3a must be satisfied. 1-2a^2=16-24a+9a^2, 11a^2-24a+15=0, which has no real solutions, so we cannot start with c=0, and we need to continue with the general case.] With P(a,b,sqrt((1-2a^2-b^2)sqrt(7)/2)), (3a+b-4)^2=(1-2a^2-b^2)sqrt(7)/2, equating the squares of the z coords for the tangential plane and the given plane. 9a^2+6a(b-4)+b^2-8b+16=(1-2a^2-b^2)sqrt(7)/2. 9a^2+6ab-24+b^2-8b+16=sqrt(7)-a^2sqrt(7)-b^2sqrt(7)/2. a^2(9+sqrt(7))+6ab+(b^2(1+sqrt(7)/2)-8b-8-sqrt(7))=0. The discriminant is 36b^2-4(9+sqrt(7))(b^2(1+sqrt(7)/2)-8b-8-sqrt(7))= 36b^2-36b^2-36b^2sqrt(7)/2+288b+288+36sqrt(7)-4sqrt(7)b^2-14b^2+32sqrt(7)b+32sqrt(7)+28= -4b^2(9sqrt(7)/2+sqrt(7)+7/2)+32b(9+sqrt(7))+316+68sqrt(7). The discriminant must be positive for real a (for example if b=0). [If b=0, P(a,0,sqrt((1-2a^2)sqrt(7)/2)) is the common point, (3a-4)^2=(1-2a^2)sqrt(7)/2. 9a^2-8a+16=(1-2a^2)sqrt(7)/2, a^2(9+sqrt(7))-8a+16-sqrt(7)/2=0. The discriminant for this is 64-2(9+sqrt(7))(32-sqrt(7)) which is negative, so b cannot be 0.] This "solution" is clearly becoming over-complicated and I suspect it's the wrong approach. Perhaps you will be able to take some of the ideas and develop them more elegantly. If I find a better method I will come back to this problem. Sorry for any inconvenience.    
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Prove, by contradiction, that if w,z∈C.... (elementary algebra)

Given the validity of the comment attached to the question, and it's resolution, we can take a look at the implication of the question. To be true for w and z generally, we can take a specific case to demonstrate the argument. Let w=0.8+0.6i, which satisfies the requirement |w|=1, a special case of |w|<1. For any value of n, the binomial expansion of (a+b)^n=1 only when a+b=1, i.e., b=1-a. The binomial expansion is (a+b)^n=a^n+nba^(n-1)+(n(n-1)/2)b^2a^(n-2)+...+b^n=1. The general term is (n!/(r!(n-r)!))a^(n-r)b^r, where 01/2. Let w=0.03+0.04i, |w|=0.05, z=1-(0.03+0.04i)=0.97-0.04i, |z|=0.9708 approx>1/2. As |w| gets smaller, |z| gets larger, so moving further away from 1/2 and towards 1. What is w+z? In all of the above cases w+z=1, by definition, and so the binomial expansion also =1. Just seen your comment. Let w=0.96+0.28i, |w|=1, z=0.04-0.28i, |z|=0.2828 approx, which is <1/2. So the proposition would not be true for the binomial expansion of (w+z)^n=1 for all n. That means we require a different expression than w+z. However, we do know that (...)^n must be 1 for all n and we need to find an alternative expression similar to the one given. Let w=a+isqrt(p^2-a^2), |w|=sqrt(a^2+p^2-a^2)=p<1. This is the general form for w. In the above, z=1-a-isqrt(p^2-a^2) and |z|=sqrt((1-a)^2+p^2-a^2)=sqrt(1-2a+a^2+p^2-a^2)=sqrt(1-2a+p^2). When p=1, this is sqrt(2(1-a)) which can be less than 1/2 (7/81/2) and wz/(w+z)=1 so ((w+z)/wz)^n=((1/z)+(1/w))^n=(wz/(w+z))^n=1. ((1/z)+(1/w))^n=(w+z)^n=1. Therefore, w and z are particular complex numbers for which the proposition is true, and the binomial expansion applies.
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Flying bearing 325 degrees 800km,turns course 235 degrees flies 950km - find bearing from A.

1. Wendy leaves airport A, flying on a bearing of 325 degrees for 800km. She then turns on a course of 235 degrees and flies for 950km. Find Wendy's bearing from A. frown 2. Wendy decides to turn and fly straight back to A at a speed of 450 km/h. How long will it take her? There are two coordinate systems used to solve this problem. The first system, used in aeronautics, is a system in which angles are measured clockwise from a line that runs from south to north. The second, used to plot graphs, measures angles counter-clockwise from the +X axis, which runs from left to right. It is necessary to convert the angles stated in the problem to equivalent angles in the rectangular coordinate system. Bearing 325 degrees is 35 degrees to the left of north. North converts to the positive Y axis, so the angle we want is the complementary angle measured up from the negative X axis. 90 - 35 = 55 If we could show a graph, we would see a line extending up to the left, from the origin at (0,0) to a point 800Km (scaled for the graph) from the origin. What we want to know is the X and Y coordinates of that point. By dropping a perpendicular line down to the X axis, we form a right-triangle, with the flight path forming the hypotenuse. We'll call the 55 degree angle at the origin angle A, and the angle at the far end of the flight path angle B. Keep in mind that angle B is the complement of angle A, so it is 35 degrees. Because we will be working with more than one triangle, let's make sure we can differentiate between the x and y sides of the triangles by including numbers with the tags. Side y1 is opposite the 55 degree angle, so... y1 / 800 = sin 55 y1 = (sin 55) * 800 y1 = 0.8191 * 800 = 655.32    We'll round that down, to 655 Km Side x1 is opposite the 35 degree angle at the top, so... x1 / 800 = sin 35 x1 = (sin 35) * 800 x1 = 0.5736 * 800 = 458.86     We'll round that up, to 459 Km At this point, the aircraft turns to a heading of 235 degrees. Due west is 270 degrees, so the new heading is 35 degrees south of a line running right to left, which is the negative X axis. Temporarily, we move the origin of the rectangular coordinate system to the point where the turn was made, and proceed as before. We draw a line 950Km down to the left, at a 35 degree angle. From the far endpoint of that line, we drop a perpendicular line to the -X axis ("drop" is the term even though we can see that the axis is above the flight path). Side y2 is opposite this triangle's 35 degree angle at the adjusted origin, so... y2 / 950 = sin 35 y2 = (sin 35) * 950 y2 = 0.5736 * 950 = 544.89    We'll round that up, to 545 Km Side x2 is opposite this triangle's 55 degree angle, so... x2 / 950 = sin 55 x2 = (sin 55) * 950 x2 = 0.8191 * 950 = 778.19    We'll round that down, to 778 Km The problem asks for the bearing to that second endpoint from the beginning point, which is where we set the first origin. We now draw our third triangle with a hypotenuse from the origin to the second endpoint and its own x and y legs. Because the second flight continued going further out on the negative X axis, we can add the two x values we calculated above. x3 = x1 + x2 = 459 Km + 778 Km = 1237 Km The first leg of the flight was in a northerly direction, but the second leg was in a southerly direction, meaning that the final endpoint was closer to our -X axis. For that reason, it is necessary to subtract the second y value from the first y value to obtain the y coordinate for the triangle we are constructing. y3 = y1 - y2 = 655 Km - 545 Km = 110 Km Using x3 and y3, we can determine the angle (let's call it angle D) at the origin by finding the tangent. tan D = y3 / x3 = 110 / 1237 = 0.0889 Feeding that value into the inverse tangent function, we find the angle that it defines. tan^-1 0.0889 = 5.08 degrees The angle we found is based on the rectangular coordinate system. We need to convert that to the corresponding bearing that was asked for in the problem. We know that the second endpoint is still to the left of the origin. The -X axis represents due west, or 270 degrees. We know that the endpoint is above the -X axis, so we must increment the bearing by the size of the angle we calculated. Bearing = 270 + 5.08    approximately 275 degrees The second part of the problem asks how long it will take Wendy (the pilot) to fly straight back to the origin. Distance = sqrt (x3^2 + y3^2) = sqrt (778^2 + 1237^2) = sqrt (605284 + 1530169)              = sqrt (2135453) = 1461.319    We'll round that one, too   1461 Km Wendy will fly 1461 Km at 450 Km/hr. How long will that take? t = d / s = 1461Km / (450Km/hr) = 3.25 hours   << 3hrs 15 mins
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x/1+x<log(1+x)<x solve using rolles mean value theorem

The function f(t)=log(1+t) is defined over [0,x] and differentiable over ]0,x[, so by the mean value theorem there exists c in ]0,x[ such that f(x)-f(0)=(x-0)f'(c), that is log(1+x)=x/(1+c). 00 it follows that x/(1+x) Read More: ...

Find Intersections of Trig Functions with different periods

When the graphs intersect f(x)=g(x) so 600sin(2(pi)/3(x-0.25))+1000=600sin(2(pi)/7(x))+500. 600(sin(2(pi)/3(x-0.25))-sin(2(pi)/7(x)))+500=0. The 7 fundamental values of x satisfying this equation are to be found at the end of this answer. Trig identities: sin(p)=sin(p+2n(pi)) and cos(p)=cos(p+2m(pi)), where n and m are integers, so sin(2(pi)(x-0.25)/3)=sin(2(pi)(x-0.25)/3+2n(pi))=sin(2(pi)(x+3n-0.25)/3) and cos(2(pi)x/7)=cos(2(pi)x/7+2m(pi))=cos(2(pi)(x+7m)/7). For g(x) the value of the function repeats for x, x+7, x+14, etc.; and for f(x) it's x, x+3, x+6 etc. The repetition of intersections of f(x) and g(x) occur for x, x+21, x+42, etc., where x is a solution of the combined equation determining the intersection points. sin(A-B)=sinAcosB-cosAsinB and sin(A+B)=sinAcosB+cosAsinB. sin(A+B)-sin(A-B)=2cosAsinB. If X=A+B and Y=A-B, X+Y=2A so A=(1/2)(X+Y) and X-Y=2B so B=(1/2)(X-Y) sinX-sinY=2cos((X+Y)/2)sin((X-Y)/2). Using these identities X=2(pi)/3(x-0.25), Y=2(pi)x/7. (X+Y)/2=(pi)((x-0.25)/3+x/7)=(pi)(7x-1.75+3x)/21=(pi)(10x-1.75)/21 (X-Y)/2=(pi)((x-0.25)/3-x/7)=(pi)(7x-1.75-3x)/21=(pi)(4x-1.75)/21 600(sinX-sinY)+500=0 so sinX-sinY=-5/6=2cos((pi)(10x-1.75)/21)sin((pi)(4x-1.75)/21) cos((pi)(10x-1.75)/21)sin((pi)(4x-1.75)/21)=-5/12. Solutions for x: 1.67126, 3.03768, 7.82582, 9.27714, 14.13103, 15.28939, 16.90941, 22.67126,... There are 7 fundamental values for x and a series can be built on each one by adding (or subtracting) 21 (LCM of 3 and 7). The intersections of the sine waves repeat the earlier pattern indefinitely. Addendum The table below shows some points on the graphs of f(x) and g(x) for the purposes of comparison and to show roughly when  f(x) intersects g(x). The comparison column f(x)~g(x) shows whether f(x) is less than or greater than g(x). Where there is a change from < to >, or vice versa, an intersection point exists between the values of x listed. The table shows two whole cycles of f(x) (0 0.25 1000 633.5 > 0.5 1300 760.3 > 0.75 1519.6 874.1 > 1 1600 969.1 > 1.25 1519.6 1040.6 > 1.5 1300 1085 > 1.75 1000 1100 < 2 700 1085 < 2.25 480.6 1040.6 < 2.5 400 969.1 < 2.75 480.6 874.1 < 3 700 760.3 < 3.5 1300 500 > 4 1600 239.7 > 4.5 1300 30.9 > 5 700 -85 > 5.5 400 -85 > 6 700 30.9 > 6.5 1300 239.7 > 7 1600 500 >  
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How are the locations of vertical asymptotes and holes different, and what role do limits play?

To talk about asymptotes and holes, you need pictures. These pictures are graphs of functions. The simplest function containing vertical and horizontal asymptotes is y=1/x, where x is the horizontal axis and y the vertical axis. The vertical asymptote is in fact the y axis, because the graph has no values that would quite plot onto the y axis, although the curve for 1/x gets very, very close. The reason is that the y axis represents x=0, and you can't evaluate 1/x when x=0. You' d have to extend the y axis to infinity both positively and negatively. You can see this if you put a small positive or negative value for x into the function. If x=1/100 or 0.01, y becomes 100. If x=-1/100 or -0.01, y becomes -100. If the magnitude of x decreases further y increases further. That's the vertical asymptote. It represents the inachievable. What about the horizontal asymptote? The same graph has a horizontal asymptote. As x gets larger and larger in magnitude, positively or negatively, the fraction 1/x gets smaller and smaller. This means that the curve gets closer and closer to the x axis, but can never quite touch it. So, like the y axis, the axis extends to infinity at both ends. What does the graph look like? Take two pieces of thick wire that can be bent. The graph comes in two pieces. Bend each piece of wire into a right angle like an L. Because the wire is thick it won't bend into a sharp right angle but will form a curved angle. Bend the arms of the L out a bit more so that they diverge a little. Your two pieces of wire represent the curve(s) of the function. The two axes divide your paper into four squares. Put one wire into the top right square and the other into the bottom left and you get a picture of the graph, but make sure neither piece of wire actually touches either axis, because both axes are asymptotes. The horizontal axis represents the value of x needed to make y zero, the inverse function x=1/y. Hence the symmetry of the graph. Any function in which an expression involving a variable is in the denominator of a fraction potentially generates a vertical asymptote if that expression can ever be zero. If the same expression can become very large for large magnitude values of the variable, potentially we would have horizontal asymptotes. I use the word "potentially" because there's also the possibility of holes under special circumstances. Asymptotes and holes are both no-go zones, but holes represent singularities and they're different from asymptotes. Take the function y=x/x. It's a very trivial example but it should illustrate what a hole is. Like 1/x, we can't evaluate when x=0. However, you might think you can just say y=1 for all values of x, since x divides into x, cancelling out the fraction. That's a horizontal line passing through the y axis at y=1. Yes, it is such a line except where x=0. We mustn't forget the original function x/x. So where the line crosses the y axis there's a hole, a very tiny hole with no dimensions, a singularity. So a hole can occur when the numerator and denominator contain a common factor. If this common factor can be zero for a particular value of x, then a hole is inevitable. Effectively it's an example of the graphical result of dividing zero by zero. With functions we can't simply cancel common factors as we normally do in arithmetic. Asymptotes and holes are examples of limits. Asymptotes can show where functions converge to a particular value without ever reaching it. Asymptotes can be slanted, they don't have to be horizontal or vertical, and they can be displaced from both axes. Graphs can aid in the solution of mathematical and physics problems and can reveal where limitations and limits exist for complicated and complex functions. Knowing where the limits are by inspection of functions also aids in drawing the graph. This helps in problems where the student may be asked to draw a graph to show the key features without plotting it formally.
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how many values of c satisfy the conclusion of the mean value theorem

??????????? satisfi the konklude av the value theorem....???????????????
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