Let f be the function given by f(x)=2x^4-4x^2+1...the questions are below, 3 part question i need help!!!?

Question

a) find an equation of the line tangent to the graph at (2,17)
b) Find the x and y coordinates of the relative maxima and relative minima. Verify your answer
c) Find the x and y coordinates of the points of inflection. Verify your answer

THANKS!!!!

Answer 1

f(x) = 2x^4 - 4x^2 + 1

a)find an equation of the line tangent to the graph at (2,17)
f'(x) = 8x^3 - 8x
f'(2) = (8)2^3 - 8(2) = 56

b) relative maxima and minima:
8x^3 - 8x = 0
x(x^2 - 1) = 0
x(x + 1)(x - 1)=0
x = 0 or x=1 or x=-1
use the 2nd derivative test:
f''(x)=24x^2 -8
f''(0)=-8 <0 so it is a rel. max
f''(1) = 24-8>0 so 1 is a rel min
f''(-1)=24-8>0 so -1 is a rel min.

f(-1) = 2(-1)^4 - 4(-1)^2 + 1 = -1
f(0) = 2*0 - 4*0 + 1 = 1
f(1) = 2(1)^4 - 4(1)^2 + 1 = -1

(-1,-1) rel min, (0,1) is a rel max and (1,-1) is a rel min.

c) points of inflection:
24x^2 - 8 = 0
3x^2 - 1 = 0
x^2 - 1/3 = 0
x = ± 1/sqrt(3)
f(-1/sqrt(3)) = 2(-1/sqrt(3))^4 - 4(-1/sqrt(3))^2 + 1
= 2(1/9) - 4(1/3) + 1 = 2/9 - 12/9 + 9/9 = - 1/9

f(1/sqrt(3)) = 2(1/sqrt(3))^4 - 4(1/sqrt(3))^2 + 1 = - 1/9

Inflection points are: (-1/sqrt(3), -1/9), (1/sqrt(3), -1/9).

Answer 2

First see if the function factors. Then find the graph of it. Then put the point for (2,17) on the graph. Find the equation of the tangent by using the formula for finding a perpendicular line. (Sorry, I haven't learned how to do maxima and minima yet or inflection).

To find the perpendicular line, just find the slope so the original slope multiplied by the new slope = -1. (The negative reciprocal). and solve for x or y from there.

Answer 3

f(x) = 2x^4 - 4x^2 + 1

a)
f'(x) = 8x^3 - 8x
f'(2) = 8*2^3 - 8*2 = 56

b) relative maxima and minima:
8x^3 - 8x = 0
x(x^2 - 1) = 0
x(x + 1)(x - 1)
x = {-1, 0, 1}
f(-1) = 2(-1)^4 - 4(-1)^2 + 1 = -1
f(0) = 2*0 - 4*0 + 1 = 1
f(1) = 2(1)^4 - 4(1)^2 + 1 = -1
relative maximum @ (0,1)
minima @ (-1,-1), (1,-1)

c) points of inflection:
24x^2 - 8 = 0
3x^2 - 1 = 0
x^2 - 1/3 = 0
x = ± 1/√3
f(-1/√3) = 2(-1/√3)^4 - 4(-1/√3)^2 + 1
f(-1/√3) = 2(1/9) - 4(1/3) + 1
f(-1/√3) = 2/9 - 12/9 + 9/9
f(-1/√3) = - 1/9

f(1/√3) = 2(1/√3)^4 - 4(1/√3)^2 + 1
f(1/√3) = - 1/9

Inflection points @ (-1/√3, -1/9), (1/√3, -1/9)

Answer 4

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