Question 5
You MAY NOT use a calculator to solve this question.
Let ƒ be the function given by ƒ(x) = 2x4 - 4x2 + 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 f(x).
f(x) = 2x4 - 4x2 + 1
f'(x) = 8x3 - 8x
y' = 8x(x2 - 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.
f'(x) = 8x(x2 - 1)
0 = 8x(x2 - 1)
8x = 0 , x2 - 1 = 0
x = 0 , x = -1 , 1 (roots of the derivative)
y- coordinates
f(x) = 2x4 - 4x2 + 1 f(1) = 2(1)4 - 4(1)2 + 1 f(0) = 2(0)4 - 4(0)2 + 1
f(-1) = 2(-1)4 - 4(-1)2 + 1 f(1) = 2 - 4 + 1 f(0) = 0 - 0 + 1
f(-1) = 2 - 4 + 1 f(1) = -1 f(0) = 1
f(-1) = -1
Since the graph has 4 as its highest power, we can assume that the graph has 4 roots. Therefore, the graph will look like
a "W". Using the number line of the derivative funcion,
- + - +
f'(x) + + +
-1 0 1
Therefore: min(-1,-1)
min (1,-1)
max (0,1)
(c) To find the inflection points, we need to find the minimum and maximum of the derivative function.
Whenever the derivative function has a root, the second derivative has a minimum or maximum. Whenever the
2nd derivative has a min or max, he parent function has an inflection point because of the change from + to -
or vice versa. It identifies its concavity.
f'(x) = 8x3 - 8x
f"(x) = 24x2 - 8
Now that we have the 2nd derivative, we can find its minimum and maximum.
f"(x) = 24x2 - 8
0 = 8(3x2 - 1)
0 = 3x2 - 1
1 = 3x2
1/3 = x2
+/- (1/3)1/2 = x (roots of the 2nd derivative)
Now that we have the roots of the 2nd derivative, we can use it to identify the inflection points.
+ - +
f"(x) + +
-(1/3)1/2 (1/3)1/2
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.
Now that we know that, we can find the y coordinates of the inflection points.
f(x) = 2x4 - 4x2 + 1
f( -1/31/2) = 2(-1/31/2)4 - 4(-1/31/2) + 1
f( -1/31/2) = 2(1/9) - 4(-1/31/2) + 1
f( -1/31/2) = 2/9 + 2(31/2)/3 + 1
f( -1/31/2) = 2 + 6(31/2) + 9
9
-being solved by jann =D
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