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# given that f(x)=(3x^2)-5, how do I find (f(x+h)-f(x))/h and (f(w)-f(x))/(w-x)?

I think that my first step would be to decompose f(x) into h(x) and w(x), but I'm not sure what to do after that.

## Research, Knowledge and Information :

### Function Notation: Evaluating at an Expression | Purplemath

For h(w) = w 2 – 3, find h ... Given that f (x) = 3x 2 + 2x, find ... You can use the Mathway widget below to practice evaluating functions at variable expressions.

### Operations on Functions | Purplemath

Performing these operations on functions is no more complicated ... and h(x) = 5 – x 3, find (f + g)(2), (h – g)(2 ... Given f (x) = 3x 2 – x + 4, find the ...

### Given that f(x)=(3x^2)-5, how do I find (f(x+h)-f... - OpenStudy

Given that f(x)=(3x^2)-5, how do I find (f(x+h)-f(x))/h and (f(w)-f(x))/ ... Given that f(x)=(3x^2)-5, how do I find (f(x+h)-f(x))/h and (f(w)-f(x))/(w-x)? Mathematics

### f(x) 2 h NOTE - California State University, Northridge

f(x+h) – f(x) 10 1. For f(x) = x 2 + 2x – 5, find ))))) ... f(x) = 2 ( x2 + 3x ) ... ( x2 – 2x + 2) = x4 – 2x3 + 2x2 (Note: The given specifications do not ...

### Evaluating Functions - Math is Fun

Evaluating Functions ... Evaluate For a Given Value: ... Example: h(x) = 3x 2 + ax − 1, You are told that h(3) = 8, can you work out what "a" is?

### THE DIFFERENCE QUOTIENT

C. Given f(x)=+3x2 = f(xh)f(x) h +− 3(xh)23x2 h ++−+ = 3(xh)23x23(xh)23x2 h3(x)23x2 ++−+ ++++ ++ = [3(xh)2](3x2) =

### Given g(x) = -3x +1, f(x) = x^2 +7, h(x) = 12/x, how do you ...

SOCRATIC ... Given #g(x) = -3x +1#, #f(x) = x^2 +7#, #h(x) = 12/x#, how do you find g ... How do you find #(h-g)(n)# given #h(n)=4n+5# and #g(n) ...

### How to find $f(\frac{1}{f(2)})$ given [math]f(x)=x ...

We find f(1/f(2)) given f(x) =x^5+x^3+x as follows: First, finding f(2): f(x) =x^5+x^3+x f(2) = 2^5 + 2^3 + 2 = 32 + 8 + 2 = 42

### Composite Functions

Composite Functions ... – 4x + 2 and g(x) = 3x – 7, find (f o g)(x). ...

## Suggested Questions And Answer :

### how do you find the equation of a line given a point and what it is perpendicular too?

how do you find the equation of a line given a point and what it is perpendicular too? i have a point and an equation in slope intercept form and i have to find the equation thats perpendicular to the eqution given . How do i do that? The equation of a line that is perpendicular to the given line has a slope that is the negative inverse of the given slope. E.G., if the given equation were y = 1/2 x + 7, the new equation would be y = -2x + b, understanding that b is the y-intercept of the new equation. Now, take the x and y values of the point you have been given and plug them into the new equation to find the value of b. If the point were (7, -10), you would have -10 = -2(7) + b. Solve that and you would find that b = 4. So, in this case, the equation would be y = -2x + 4. Follow the same steps, using the values from your equation and point. You will end up with the correct equation for your problem.

### how do you get an equation of a tangent line to a parabola given the point of tangency

You need to find the derivative dy/dx of the curve and substitute x and y, if necessary, to find the slope at a particular given point. The slope is the tangent at the point. The standard linear equation y=ax+b can then be constructed, because the tangent will be a. To find b, the y intercept, substitute the coords of the given point into x and y and you'll be able to find b. That gives you the line. For example, let y=px^2+qx+r be the equation of a parabola, where p, q and r are constants, then dy/dx=2px+q. Let's say you're given the point on the curve where x=A, then dy/dx=2pA+q. This is a number because p, q and A are all known values (you'll be told what they are). This is your slope for the tangent, so a=2pA+q, and y=(2pA+q)x+b is the equation of the tangent line. To find b we first find out what y is on the curve when x=A. We know y=pA^2+qA+r. To make it easier to read, call this y coord B. So we substitute (A,B) in our linear equation to find b=B-A(2pA+q). It's much easier when you have actual numbers rather than symbols. In the same way you can find the equation of the perpendicular because its slope is related to the slope of the tangent. It's -1/(2pA+q), the negative reciprocal of the tangent. I hope this helps your understanding.

### by using inclusion exclusion principle and derangements find the solution?

QUESTION 1 First the Venn diagram:   KEY A Females/outside 19-20 age group/not involved in sports or recreation activities B Males only C 19-20 year old females only  D Involved in Sports/Recreation (S/R) only E Males involved in S/R, but not 19-20 years old F 19-20 year olds involved in S/R G 19-20 year old males not involved in S/R H 19-20 year old males involved in S/R A+B+C+D+E+F+G+H=1400 (all employees) B+E+G+H=675 (males) C+F+G+H=682 (19-20 year olds) D+E+F+H=684 (involved in S/R) G+H=195 (19-20 year old males) E+H=467 (males in S/R) F+H=318 (in S/R) H=165 (19-20 year old males in S/R) (a) Requires us to find A. Given H=165 we can find F=153, E=302, G=30. Given E=302, F+H=318, D=64. Given F=153, G+H=195, C=334. Given E=302, G+H=195, B=178. Now we can find A=1400-(178+334+64+302+153+195)=174. (b) To find G; G=30. (c) To find C; C=334. QUESTION 2 (A) First 5 integers 1 2 3 4 5. There are 24 ways of arranging 1 2 4 5 including the numbers being in their natural position, so we need to exclude permutations where that would happen. 24 permutations: 1245 1254 1425 1452 1524 1542  2145 2154 2415 2451 2514 2541 4125 4152 4215 4251 4512 4521 5124 5142 5214 5241 5412 5421  9 remain not crossed out, so there are 9 permutations with only 3 in its natural position. (b) There are 8: 41325, 51342, 24315, 52314, 14352, 42351, 15324, 25341

### how do you find the slope of a line sagment?

how do you find the slope of a line sagment? you are given two points which define a line segment. Find the slope of the line sagment, and then use this slope to write the equation of a parallel line [and a] perpendicular line which pass through the new y-intercept 1. line segment (1,8) and (7,-4)  y-intercept of new lines: (0,-7) The slope is (y2 - y1) / (x2 - x1) m = (-4 - 8) / (7 - 1) = -12/6 = -2 This line segment is y = -2x + b. We don't know the y-intercept, but we are not concerned with that. A parallel line has the same slope but a different y-intercept. We are given the y-intercept, so the equation is y = -2x - 7. A line perpendicular to the given line segment has a slope that is the negative reciprocal: 1/2. The equation of that line is y = (1/2)x - 7.

### A standard Hyperbola has its vertices ar 5,7 and 3,-1.Find its equation and its asymptotes

this equation expands toThe vertices share no common coordinate so the hyperbola cannot be orientated in the usual x-y form. The hyperbola must therefore be skew (tilted axis) and the line joining the vertices is sloping. This line is given by the equation y=((7+1)/(5-3))x+c where c is to be found. y=4x+c, and -1=12+c, plugging in (3,-1). c=-13. Therefore the axis of the hyperbola is y=4x-13. The centre or origin of the hyperbola lies midway between them at ((1/2)(5+3),(1/2)(7-1))=(4,3) and the equation of the conjugate axis is given by y=-x/4+k where k is found by plugging in the midpoint: 3=-4/4+k, so k=4 and y=4-x/4 is the equation of the conjugate axis. Let's define the axes of the hyperbola as X and Y instead of x and y. We can place the centre of the hyperbola at (0,0) in the X-Y plane so that in this plane the equation of the hyperbola is Y^2/A^2-X^2/B^2=1. The vertices are found by finding Y when X=0, so (0,A) and (0,-A) are the vertices. The asymptotes are given by (Y/A+X/B)(Y/A-X/B)=0. So Y=AX/B and Y=-AX/B are their equations. The distance between the vertices is 2A so we can work this out from the x-y coordinates of the vertices using Pythagoras' Theorem: 2A=sqrt((7+1)^2+(5-3)^2)=sqrt(68). A=sqrt(68)/2, A^2=68/4=17. Since the hyperbola is rectangular, B=A, and the equation of the hyperbola is Y^2-X^2=17. The asymptotes then become Y=X and Y=-X. Now we have to transform X-Y to x-y, so that the equation can be expressed in the x-y plane. We use geometry and trigonometry  for this. Let O'(4,3) be the location of the origin of the X-Y plane. Let P(x,y) be any point. In the X-Y plane this is P'(X,Y). Join O'P'=O'P. This line can be represented in both reference frames. O'P^2=X^2+Y^2=(x-4)^2+(y-3)^2 by Pythagoras. The gradient of the Y axis with respect to the x-y frame is 4, and we can represent this as tan(w) so w=tan^-1(4) or tan(w)=4; sin(w)=4/sqrt(17); cos(w)=1/sqrt(17). The angle z that O'P makes with the line y=3 is given by tan(z)=(y-3)/(x-4). sin(z)=(y-3)/O'P; cos(z)=(x-4)/O'P. Let v=90-w. sin(v+z)=Y/O'P; cos(v+z)=X/O'P. v+z=90-w+z=90-(w-z), so sin(v+z)=cos(w-z)=cos(w)cos(z)+sin(w)sin(z)=(x-4)/(O'Psqrt(17))+4(y-3)/(O'Psqrt(17))=(x+4y-16)/(O'Psqrt(17)). cos(v+z)=sin(w-z)=sin(w)cos(z)-cos(w)sin(z)=4(x-4)/(O'Psqrt(17))-(y-3)/(O'Psqrt(17))=(4x-y-13)/(O'Psqrt(17)). Y=O'P*sin(v+z)=(x+4y-16)/sqrt(17); X=O'P*cos(v+z)=(4x-y-13)/sqrt(17). So P maps to the x-y plane using these transformations. Y^2-X^2=17 becomes (x+4y-16)^2-(4x-y-13)^2=289, the equation of the hyperbola. This expands to x^2+8x(y-4)+16y^2-128y+256-(16x^2-8x(y+13)+y^2+26y+169)=289; -15x^2+16xy+15y^2+72x-154y-202=0. The asymptotes are x+4y-16+4x-y-13=0; 5x+3y-29=0 and x+4y-16-4x+y+13=0; 5y-3x-3=0.

### Given a ratio of 2:3 find the volume of the smaller of the larger cylinder is 252pi ft squared.

Question: Given a ratio of 2:3 find the volume of the smaller of the larger cylinder is 252pi ft squared. The volumes of teh two cylinders is in the ratio of 2:3. A ratio of 2:3 is the same as a fraction of 2/3. If teh two volumes are V1 and V2 (whwre V1 is teh smaller volume) then we can write V1:V2 = 2:3 or, we can write (using fractions) V1/V2 = 2/3, or V1 = (2/3)*V2 We are given V2 = 252π ft squared, then V1 = (2/3)*252π = 168π Volume of smaller cylinder is 168π ft squared

### Find a normal vector of the curve 6x^2+2y^2+z^2 = 225 at the given point P : (5; 5; 5).

Find a normal vector of the curve 6x^2+2y^2+z^2 = 225 at the given point P : (5; 5; 5).​ The normal vector at a point on a surface is given by (1) where and are partial derivatives. and (x0, y0) = (5,5) we have z = (225 - 6x^2 - 2y^2)^(1/2) = f(x,y), giving us fx = -6x(225 - 6x^2 - 2y^2)^(-1/2),     and   fx(5,5) = -30(225 - 150 - 25)^(-1/2) = -30(25)^(-1/2) = -6 fy = -2y(225 - 6x^2 - 2y^2)^(-1/2),     and   fx(5,5) = -10(225 - 150 - 25)^(-1/2) = -10(25)^(-1/2) = -2 Therefore, N = | -6 |                         | -2 |                         | -1 |

### Use the given zero to find the remaining zeros. h(x)=3x^4+10x^3+19x^2+90x-73 zero:-3i

Use the given zero to find the remaining zeros. h(x)=3x^4+10x^3+19x^2+90x-73 zero:-3i There’s an error in your quartic. It should be: h(x)=3x^4+10x^3+19x^2+90x-72 You are given that one of the zeros is -3i. i.e. one of the factors is (x + 3i) If a polynomial has a complex factor, or root, then it must have a 2nd root which is a complex conjugate of the 1st root. i.e. a 2nd factor is (x – 3i). Since both (x + 3i) and (x – 3i) are factors of h(x) then so also is their product, Which is (x + 3i)(x – 3i) = x^2 + 9 Which means that (x^2 + 9) divides into (3x^4+10x^3+19x^2+90x-72). The result of this division is: 3x^2 + 10x – 8 I.e. (x^2 + 9)(3x^2 + 10x – 8) = 3x^4+10x^3+19x^2+90x-72 The factor (3x^2 + 10x – 8) factorises as (3x – 2)(x + 4) Giving roots: x = 2/3 and x = -4. The four roots are: x = 3i, x = -3i, x = 2/3, x = -4

### line d y=2x+4 ,e y=2x+c find c and the lines are parallel and have 6 units between them

line d y=2x+4 ,e y=2x+c find c and the lines are parallel and have 6 units between them There are two lines that satisfy the given parameters, one above and to the left of line d, and one below and to the right of line d. To find a point 6 units away from line d, we need a line that is perpendicular to d. We get the equation of that line by using the negative reciprocal of the m component, 2: that gives us -1/2. The equation of this line is y = -1/2 x + 4, using the same y intercept. A segment of that line that is 6 units represents the hypotenuse of a right triangle. If we are looking at the segment that extends to the left of the given y intercept, we consider the distance to be negative (-6). If we are looking at the segment that extends to the right of the given y intercept, we consider the distance to be positive (+6). To find the x and y co-ordinates, we use the sine and the cosine of the angle (we'll call it a) formed by this new line and the x axis. The slope m of a line is the y distance divided by the x distance. That is also the definition of the tangent of the angle. Using the inverse tangent, we determine that the angle is tan^-1 (-1/2) =  -26.565 degrees. We need to keep all the signs straight in order to get the correct values. The sine of -26.565 degrees is -0.4472. The cosine of -26.565 degrees is 0.8944. y1 / -6 = sin a y1 = -6 sin a = -6 * -0.4472 = 2.6832 This is measured from a horizontal line through the y intercept, because we are constructing a right-triangle with one corner at the y intercept, so the point's y co-ordinate is actually y = 2.6832 + 4 = 6.6832 x / -6 = cos a x = -6 * 0.8944 = -5.3664 Making a quick check using x^2 + y^2 = r^2: (-5.3664)^2 + 2.6832^2 = 28.79825 + 7.19956 = 35.99781   -6^2 = 36 Close enough considering the rounding in the calculations. We have co-ordinate (-5.3664, 6.6832) lying 6 units from line d. Substituting those values into y = 2x + c we have 6.6832 = 2 * (-5.3664) + c 6.6832 = -10.7328 + c 6.6832 + 10.7328 = c = 17.416 The equation of a line 6 units away from d and parallel to it, located above it, is y = 2x + 17.416    <<<<<<<< Working in the other direction, on the line below and to the right of line d, y2 / 6 = sin a y2 = 6 * sin a = 6 * -0.4472 = -2.6832 (that is, 2.6832 below d's y intercept) y = 4 - 2.6832 = 1.3168 x / 6 = cos a x = 6 * cos a = 6 * 0.8944 = 5.3664 The co-ordinates for this point, (5.3664, 1.3168), when substituted into the equation y = 2x + f (f is a new y intercept) gives 1.3168 = 2 * 5.3664 + f 1.3168 - 10.7328 = f = -9.416 The equation of a line 6 units away from d and parallel to it, located below it, is y = 2x - 9.416    <<<<<<<<

### Given P1 = (a,b) and P2 = (c,d), find the distance between the two points.

Given P1 = (a,b)  and  P2 = (c,d), find the distance between the two points. Please explain you to solve it, thank you. Form a rigth-triangle, with the distance between the two points represented by the hypotenuse. The length of the x side is (c - a).  The length of the y side is (d - b). The Pythagorean theorem tells us that d^2 = x^2 + y^2, or d^2 = (c - a)^2 + (d - b)^2 Taking the square root of both sides of the equation gives us d = √((c - a)^2 + (d - b)^2) For our purposes here, there is no need to multiply the quantities on the right side of the equation. That would result in an unnecessarily complex expression. If you were given actual values for a, b, c and d, you would perform the subtractions, square the numbers you get, add the squares, then take the square root.