`f(x) = (x^4)e^(-x)` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the...

`f(x) = (x^4)e^(-x)` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the inflection points.

Asked on by enotes

Textbook Question

Chapter 4, 4.3 - Problem 18 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

1 Answer | Add Yours

shumbm's profile pic

Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

f is defined everywhere and is infinite differentiable.

(a) `f'(x) = e^(-x)*(4x^3 - x^4) = x^3*e^(-x)*(4-x).`

It is positive for x on (0, 4) and f is increasing there, and negative on `(-oo, 0)` and on `(4, +oo)` , f is decreasing there.

(b) therefore f has a minimum at x=0, f(0)=0, and naximum at x=4, `f(4)=(4/e)^4 approx4,69.`

(c) `f''(x) = e^(-x)*(12x^2 - 4x^3 - (4x^3 - x^4)) = e^(-x)*(x^4 - 8x^3 + 12x^2) =` `x^2*e^(-x)*(x^2 - 8x + 12) = x^2*e^(-x)*(x-2)*(x-6).`

It is negative for x on (2, 6) where f is concave downward, and positive for x on (-oo, 2) and (6, +oo), f is concave upward there.

2 and 6 are inflection points, 0 is not (f'' doesn't change sign).

We’ve answered 318,930 questions. We can answer yours, too.

Ask a question