`f(x) = x^4 - 4x^3 + 2` Find all relative extrema, use the second derivative test where applicable.

Textbook Question

Chapter 3, 3.4 - Problem 35 - Calculus of a Single Variable (10th Edition, Ron Larson).
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thilina-g | College Teacher | (Level 1) Educator

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The first derivative,

`f'(x) = 4*x^3-4*3*x^2+0 = 4x^3-12x^2=4(x^3-3x^2)`

At critical points, `f'(x) = 0`

`4(x^3-3x^2) = 0`

`x^2(x-3) = 0`

This gives, `x = 0 or x = 3`

At `x = 0, f(x) = 2` and,

at `x=3, f(x) = 3^4-4*3^3+2 = 81-108+2=-25`

Therefore, the points` (0,2) ` and `(3,-25)` are critical points.

Second derivative test

`f''(x) = 4*(3x^2-3*2*x) = 4(3x^2-6x) = 12x(x-2)`

at `x = 0, f"(x) = 0` . Therefore, we have an inflection point at `(0,2).`

at `x=3, f"(x) = 12*3(3-2) = 36 gt0` , therefore, we have an local minimum at `(3,-25).` 

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