# Question 5

## You MAY NOT use a calculator to solve this question.

Let ƒ be the function given by ƒ(x) = 2x^{4} - 4x^{2} + 1.

*(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 minima. Verify your answer.

*(c)* Find the x- and y-coordinates of the points of inflection. Verify your answer.

**Solution**

*(a)* In order to find the equation of the tangent line, we need to find the derivative of ƒ(x).

ƒ(x) = 2x^{4} - 4x^{2} + 1

ƒ'(x) = 8x^{3} - 8x

y' = 8x(x^{2 }- 1)

y'(-2) = 8(-2)[(4) - 1]

y'(-2) = -16[3]

y'(-2) = -48 (slope of the tangent line)

Point-Slope Formula:

y - y1 = m(x - x1)

y - (17) = -48(x - (-2))

y - 17 = -48(x + 2)

y = -48x - 96 + 17

y = -48x - 79 (equation of the tangent line)

*(b)* Now that we have the derivative of f(x), we can use it to find the minimum and maximum of the parent function.

If we can recall, whenever the parent function has a minimum and/or maximum, the derivative function has a root at that particular point.

ƒ'(x) = 8x(x^{2 }- 1)

0 = 8x(x^{2 }- 1)

8x = 0 , x^{2 }- 1 = 0

x = 0 , x = -1 , 1 (roots of the derivative)

y- coordinates

ƒ(x) = 2x^{4} - 4x^{2} + 1 ƒ(1) = 2(1)^{4} - 4(1)^{2} + 1 ƒ(0) = 2(0)^{4} - 4(0)^{2} + 1

ƒ(-1) = 2(-1)^{4} - 4(-1)^{2} + 1 ƒ(1) = 2 - 4 + 1 ƒ(0) = 0 - 0 + 1

ƒ(-1) = 2 - 4 + 1 ƒ(1) = -1 ƒ(0) = 1

ƒ(-1) = -1

Since the graph has a highest degree of 4, we can assume that the graph may have a max of 4 roots. Therefore, the graph will look like

a "W" or an "M" if the leading coefficient of the highest power is negative. Using the number line of the derivative function,

- + - +

ƒ'(x)~~ + + + ~~

-1 0 1

Therefore: min(-1,-1)
min (1,-1)

max (0,1)

*(c)* To find the points of inflection, we need to find the minimum and maximum of the derivative function.

Whenever the derivative function has a minimum or maximum, the second derivative has a root. And whenever the

2nd derivative has a root, the parent function has an inflection point because of the change from + to -

or vice versa. It identifies the parent graph's concavity.

ƒ(x) = 8x^{3} - 8x

ƒ"(x) = 24x^{2} - 8

Now that we have the 2nd derivative, we can find its roots.

ƒ"(x) = 24x^{2} - 8

0 = 8(3x^{2} - 1)

0 = 3x^{2} - 1

1 = 3x^{2}

1/3 = x^{2}

Now that we have the roots of the 2nd derivative, we can use it to identify the inflection points.

+ - +

ƒ"(x)~~ + + ~~

-√(1/3) √(1/3)

If there is a change from + to -, the concavity of the parent function change from concave up to concave down.

If there is a change from - to +, the concavity of the parent function change from concave down to concave up.

-solved by jann =D

^{ }

± √(1/3) = x (roots of the 2nd derivative)

## Comments (0)

You don't have permission to comment on this page.