At what point on f(x)=-3/2+6x+3x^2-(1/2)x^4

Question

does the tangent line have max slope?

Here are the answer choices:
a. (1,-5)
b. (-1,-4)
c. (-1,7)
d. (1,8)
e. None of these (don't want to pick this unless I'm absolutely sure of the answer)

 

Answer 1

The function f(x) = (-3/2) + 6x + 3x² - (1/2)x⁴

Rewrite the polynomial function y = - (1/2)x⁴ + 3x² + 6x - (3/2)

Differentiating with respect to x.

y' = - (4/2)x³ + 6x + 6

y' = - 2x³ + 6x + 6

Find the second derivative.

y'' = - 6x² + 6

Equate y'' = 0

 - 6x² + 6 = 0

6x² = 6

x² = 1

x = ± 1

Substitute the x values in first derivative to find maximum or minimum points.

For x = 1

y' = - 2(1)³ + 6(1) + 6 = 10

For x = - 1

y' = - 2(-1)³ + 6(-1) + 6 =  2

Maximum slope occurs at x = 1

Using the x - value find the corresponding y - value with the curve.

f(1) = (-3/2) + 6(1) + 3(1)² - (1/2)(1)⁴

= (-3/2) + 6 + 3 - 1/2

= 9 - 2

 = 7

The point is (1, 7)

None of these.

Option e is correct.