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# Given f(x) = [3x^3 + 17x^2 + 6x + 1] / [2x^3 - x + 3] , find the nature of the turning points.

Please the Newton's method steps are important for me when solving for dy/dx = 0.

## Research, Knowledge and Information :

### x^3-6x^2+17x Find the point of inflection of the graph of the ...

x^3-6x^2+17x Find the point of ... is a continuous function with f"(x)=-5x^2(2x-1)^2(3x+1)^3 , find the set of ... Find the maximum number of turning points ...

### Find the real solutions of the equation. -8x^3-17x^2+7x=0

... describe the nature of the solutions. -2x^2+3x-7=0 What ... (x)=2x^4-17x^3+47x^2-58x+24 (A) Find all ... 3 +44x^2 -17x -5 given that -5/2 is a zero of f(x) ...

### Pre Calculus 2014 Flashcards | Quizlet

Start studying Pre Calculus 2014 ... Find all local maxima and minima and points of inflection of the function. f(x)=-2x^3+6x^2 ... Which is the function f(x)=x^2- 6x ...

### Chapter 4: Polynomials: Solved problems - NIU

Chapter 4: Polynomials: Solved problems 1. ... let 2x 4-x 3 +x 2 +3x+1 = f(x) ... [x+3][x 2 +2x+1] = [x 3 +2x 2 +x+3x 2 +6x+3] = [x 3 +2x+3] = ...

### Algebra 2 STUDY GUIDE AII.7 AII.8 Polynomials Mrs. Grieser

... (9x3 – 18x2 – x + 2) ÷ (3x + 1) b) (3x3 – 2x2 + 4x ... = 2x4 + 6x3 – 5x2 – 60 at x = -1 3) Given f(x) ... .2, 1.5) b) 3rd degree; 2 max turning points; ...

### Find the following for the function f(x)=(x+6)^2(x-3)^2 ...

Find the following for the function f(x)=(x+6)^2 ... the maximum number of turning points on the graph of f. d) ... polynomial f(x) is given by f(x) = 2x^3 + 9x^2 ...

### 3.2 The Factor Theorem and The Remainder Theorem

3.2 The Factor Theorem and The Remainder Theorem 257 ... x 2 x3 + 4x2 5x 14 x3 2x2 6x2 5x ... form given in Theorem3.4. 1. 2 5x3 2x2 + 1 (x 3)2. x3 + 8

### Chapter 3 Polynomial and Rational Functions - Cengage

Chapter 3 Polynomial and Rational Functions ... How many turning points do you see in each function? ... to help ﬁnd all the zeros of f(x) = x4 +x3 − 6x2 − 2x+4.

## Suggested Questions And Answer :

### Determine the first derivative (dy/dx) : y= f(x)=x^3-3x+2

Please note that the title to this question has y=f(x)=x^3-3x+2 but the text of the question has y=f(x)=3x^3-3x+2, which is the equation I have used in my answer. (a) 1. f'(x)=dy/dx=9x^2-3 using the rules of differentiation 2. f"(x)=18x (gives the nature of any turning point) 3 and 4: f'=0 at a turning point, so 9x^2=3 and x^2=1/3 so x=±1/√3 or ±√3/3 in rational form. When x>0 f">0 and when x<0 f"<0. This means that x=√3/3 is a minimum and x=-√/3 is a maximum.  5. There is no point of inflexion. 6. The graph is shown below. The intercepts are y=2 and the root(s) of 3x^3-3x+2=0. They only point where the curve intercepts the x axis is about x=-1.24. If f(x)=x^3-3x+2 was the intended equation, the graph will look similar (not the same) but all the answers will not fit. Nevertheless I think you can work out the answers for yourself given the general idea of how to do so in this solution. The major difference is that x^3-3x+2 factorises: (x+2)(x-1)^2 with roots x=-2 and x=1 (x intercepts, so the curve cuts or touches the x axis twice). Also the max and min are at x=±1 (max at x=-1 and min at x=1). The graph below of x^3-3x+2 clearly shows the intercepts with the curve just touching (not intercepting) the x axis at x=1, while x=-1 is a clear intercept. The turning points are (-1,4) (max) and (1,0) (min). If you would like a full solution to this, add a comment to this answer and I will answer using a comment reply.

### f(x)=xcubed+xsquared-2x...determine x intercept and turning point

f(x)=x^3+x^2-2x=x(x^2+x-2)=x(x+2)(x-1) The y intercept is when x=0, so y=0 (origin); the x intercept is when y=0, so x(x+2)(x-1)=0, so x=0 (origin), 1 and -2. The gradient or slope is f'(x)=3x^2+2x-2 and is zero at a turning point. Using the formula to solve the quadratic: x=(-2+sqrt(4+24))/6=(-1+sqrt(7))/3=0.5486 and -1.2153 (approx). The coords of the turning points are (0.5486,-0.6311) and (-1.2153,2.1126). The nature of these turning points is given by f"(x)=6x+2, which is positive (minimum) at the first turning point and negative (maximum) at the second. The graph zigzags by starting in the negative quadrant (QIII) crossing the x axis at -2 to QII, reaching a maximum at the second turning point, passing through the origin to reach the minimum at the first turning point in QIV, and crossing the x axis again at 1 to continue rising in QI.

### What is Mike's speed given the information below

In the first part of the problem, Andrew, traveling at a speed of v1, travels 4 miles, while Mike, traveling at speed v2, travels (d - 4) miles. They leave from their respective starting points at the same time, so the time it takes for them to meet and pass is the same for both. t = d / s 1. t1 = 4/v1 = (d - 4) / v2 Multiply both sides by v1 to eliminate the denominator on the left side, and multiply both sides by v2 to eliminate the denominator on the right side. 2. (4 / v1) * v1 * v2 = ((d - 4) / v2) * v1 * v2 3. 4v2 = (d - 4) v1 Divide both sides by four to get the value of v2 4. v2 = ((d - 4)v1) / 4 In the second part of the problem, Andrew has reached Simburgh (d) and turned around, travelling another 2 miles, or (d + 2), while Mike has reached Kirkton and turned around, travelling another (d - 2) miles, for a total of d + (d - 2) = (2d - 2) miles. Again, their times are equal when they meet and pass. 5. t2 = (d + 2) / v1 = (2d - 2) / v2 As in the first part, multiply both sides by v1 to eliminate the denominator on the left side, and multiply both sides by v2 to eliminate the denominator on the right side. 6. ((d + 2) / v1) * v1 * v2 = ((2d - 2) / v2) * v1 * v2 7. (d + 2)v2 = (2d - 2)v1 Divide both sides by (d + 2) go get the value of v2 8. v2 = ((2d - 2)v1) / (d + 2) We have two equations for v2, equation 4 and equation 8. The problem states that v2 remains the same throughout the journey. Therefore: ((d - 4)v1) / 4 = ((2d - 2)v1) / (d + 2) Once again, multiply both sides by both denominators. (((d - 4)v1) / 4) * 4 * (d + 2) = (((2d - 2)v1) / (d + 2)) * 4 * (d + 2) v1 * (d - 4) * (d + 2) = v1 * (2d - 2) * 4 Divide both sides by v1, eliminating speed from this equation. (d - 4) * (d + 2) = (2d - 2) * 4 d^2 - 4d + 2d - 8 = 8d - 8 d^2 - 2d - 8 = 8d - 8 Subtract 8d from both sides and add 8 to both sides. (d^2 - 2d - 8) - 8d + 8 = (8d - 8) + 8d + 8 = 0 d^2 - 10d = 0 Factor out a d on the left side. d * (d - 10) = 0 One of those factors is equal to 0 (to give a zero answer). d = 0 doesn't work; we already know the distance is more than 4 miles. d - 10 = 0 d = 10    <<<<<   That's the answer to the first question, how far is it? We'll substitute that into equation 4 to find v2 in relation to v1. v2 = ((d - 4)v1) / 4 = ((10 - 4)v1) / 4 v2 = 6v1 / 4 = (6/4)v1 v2  = 1.5 * v1   <<<<<< That's the answer to the second question No matter what speed you choose for Andrew (v1), Mike's speed is one-and-a-half times faster. Let's set Andrew's speed to 6mph and solve equation 1. t1 = 4/v1 = (d - 4) / v2 t1 = 4mi / 6mph = (10 - 4) / (1.5 * 6mph) 4/6 hr = 6/9 hr 2/3 hr = 2/3 hr With Andrew travelling at 4 mph, and Mike travelling at 6 mph, it took both of them 2/3 of an hour to reach a point 4 miles from Kirkton.

### The temperature in a greenhouse from 7:00p.m. to 7:00a.m. is given by f (t)= 96 - 20sin (t/4), where f (t) is measured in Fahrenheit, and t is the number of hours since 7:00 p.m.

The temperature in a greenhouse from 7:00p.m. to 7:00a.m. is given by f (t)= 96 - 20sin (t/4), where f (t) is measured in Fahrenheit, and t is the number of hours since 7:00 p.m. A) What is the temperature of the greenhouse at 1:00 a.m. to the nearest Fahrenheit? (B) Find the average temperature between 7:00 p.m. and 7:00 a.m. to the nearest tenth of a degree Fahrenheit. (C) When the temperature of the greenhouse drops below 80 degreeso Fahrenheit, a heating system will automatically be turned on a maintain the temperature at minimum of 80 degrees Fahrenheit. At what values of the to the nearest tenth is the heating system turned on? (D) The cost of heating the greenhouse is \$0.25 per hour for each degree. What is the total cost to the nearest dollar to heat the greenhouse from 7:00 p.m. and 7:00 a.m.?   The equation is: f(t) = 96 – 20sin(t/4), 0 <= t <= 12 (A)  At 1.00 a.m. t = 6 f(6) = 96 – 20.sin(6/4) = 96 – 20*0.99749 = 96 – 19.9499 f(6) = 76 ⁰F (B)  The average temperature would need to be worked out by sampling the temperature at different times throughout the night. Divide the temperature range into N equal intervals, giving N+1 sampling points. We would then have T1 = f(δt), T2 = f(2δt), T3 = f(3δt), ... ,Tn = f(nδt) Where δt = range/N = 12/N, and n = 0..N Giving Tn = 96 – 20.sin((12n/N)/4) = 96 – 20.sin(3n/N) Then Tav = (1/N)*sum(Tn, n = 0 .. N) i.e. Tav = (1/N)*sum(96 – 20.sin(3n/N), n = 0 .. N Tav = 96 – 20. (1/N)*sum(sin(3n/N), n = 0 .. N I used Maple to evaluate the above summation. The results are tabulated as follows.                                  Average Temperature Num Intervals            3       4        6       10       20      50     100    200 Tav over the range 83.39 83.01 82.78 82.69 82.69 82.71 82.72 82.73 As can be seen from the table the temperature is averaging out at:  Tav = 82.7 ⁰F (C)  T = f(t) = 96 – 20sin(t/4), 0 <= t <= 12 At T = 80 ⁰F,            96 – 20sin(t/4) = 80 20.sin(t/4) = 96 – 80 = 16 sin(t/4) = 0.8 t/4 = 0.927295 t = 3.70918 t = 3.7 (to nearest tenth) (D)  The temperature will (normally) drop to 80 ⁰F after t = 3.7 hours and rise again to 80 ⁰F when t = 12 – 3.7 = 8.3 hours. Heating system is turned on for 8.3 – 3.7 = 4.6 hours Cost of heating is 4.6*80*0.25 = 4.6*20 = 92 Cost = \$92

### Calculus 1 help

Find the absolute maximum and absolute minimum values of f on the given interval. f(x)= (x^2-4)/(x^2+4), [-4,4] Maxima and minima are found where f'(x) = 0 f'(x) = 2x/(x^2 + 4) - 2x(x^2 - 4)/(x^2 + 4)^2 f'(x) = 16x/(x^2 + 4)^2 A turning point is found when x = 0, giving f'(x) = 0 This turning point is at the point (0, -1) As x -> ꝏ, x^2 –  4 -> x^2 As x -> ꝏ, x^2 + 4 -> x^2 Therefore, as x -> ꝏ, (x^2 –  4)/(x^2 + 4) -> x^2/x^2 = 1 As x -> -ꝏ, (x^2 –  4)/(x^2 + 4) -> x^2/x^2 = 1 i.e. As x -> ±ꝏ, (x^2 –  4)/(x^2 + 4) ->  1 So, the line x = 1 is a horizontal asymptote for the function f(x) = (x^2 – 4)/(x^2 + 4) Although it is not a local maximum, f(x) has a maximum value of 1 as x -> ±ꝏ The local minimum would be f(x) = -1 at x = 0

### The graph of f(x) =ax^3+bx^2 has a local turning point at (2;-4) find the value of a and b

y=ax^3+bx^2...or y=x^2*(ax+b) a "terning point" meen the slope chaenj from positiv tu negativ or the opposit for a kontinyuus kerv, that meen it gotta go thru ZERO. slope=dy/dx=3ax^2+2bx d2y/dx2=6ax+2b=0 at x=2, so 2*6a=-2b, or b=-6a yu sae kerv go thru point=(2,-4)...y=ax^3+bx^2...8a+4b=-4...4b=-8a-4...b=-2a-4 thus b=-6a=-2a-4, so 2a-6a=-4...-4a=-4...a=1 b=-6a=-6

### What's the point of intersection for y = x^1/2 and y = e^-3x in the interval (0,1)?

The point of intersection is given by x^1/2=e^-3x. It turns out that x=0.23873 approx. To find the volume of revolution, you need a picture of the graphs so the you can see what you are doing.   Integral of ydx between x=0 and 1 I'll write as S[0,1](ydx). This would give you the area enclosed. Volume of rev=(pi)S[0,1](y^2dx) the sum of the volumes of thin discs about the x axis. Up to the point of intersection, the volume is governed by y=x^1/2, and after that it's the other function. (pi)S[0,0.23873](y^2dx)=(pi)S[0,0.23873](xdx)=(pi)x^2/2[0,0.23873]=0.08952 (first part of integration) (pi)S[0.23873,1]((e^-6x)dx)=(pi)(-e^-6x)/6[0.23873,1]=(0.23873/6-e^-6/6)(pi)=0.12370 (second part of integration) Note that e^-6x=(e^-3x)^2=x=0.23873 at the intersection. Total volume=0.08952+0.12370=0.2132 approx.

### 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.

### A 22 kilos child is sitting at A(-7, -3) and a 47.5 kilos child is at B(1,4), where units are in feet

For a general solution to this problem let A be the point (x1,y1) and B be (x2,y2). Let AB=R and let a on AB be the distance between A and the fulcrum F(x,y). The masses are M1 at A and M2 at B. The moments about the fulcrum must be equal for equilibrium, so M1a=M2(R-a) and M1a+M2a=M2R, making a/R=M2/(M1+M2). By similar triangles (y-y1)/(y2-y1)=(x-x1)/(x2-x1)=a/R=M2/(M1+M2). Also, R=√(y2-y1)^2+(x2-x1)^2) and a=√(y-y1)^2+(x-x1)^2). Therefore (x-x1)(M1+M2)=(x2-x1)M2; x(M1+M2)-x1M1-x1M2=x2M2-x1M2. So x=(x1M1+x2M2)/(M1+M2) and y=(y1M1+y2M2)/(M1+M2). Therefore the coordinates of F are ((x1M1+x2M2)/(M1+M2),(y1M1+y2M2)/(M1+M2)). These formulae can be used in all problems of this nature. We have M1=22kg and M2=47.5kg, so M1+M2=69.5kg. x1=-7, y1=-3, x2=1, y2=4, so F is ((-7*22+47.5)/69.5,(-3*22+4*47.5)/69.5)=(-213/139,248/139)=(-1.53',1.78'). Note that the formula can be used to find any of the values M1, M2, x1, x2, x, y1, y2, y given the coordinates of A and B or F or M1 or M2.