On what interval is the function f (x) = x^3 e^x increasing?

2 Answers
Oct 23, 2017

f(x) is increasing in (-3,0), (0, oo).

Explanation:

STEP 1: Remember that our function, f(x), is increasing when the derivative, f'(x) > 0. So our first step is to find f'(x).

f'(x) = 3x^2 e^x + x^3 e^x => x^2 e^x (3 + x)

STEP 2: Find the zeros of the function.

x = 0, x = -3

STEP 3: Use a sign chart to determine when f'(x) > 0 at each interval.

STEP 3a: Plug any value less than -3 into the f'(x).

When x < -3, f'(x) < 0

STEP 3b: Plug in any value between -3 and 0 into f'(x).

When -3 < x < 0, f'(x) > 0

STEP 3c: Plug any value greater than 0 into the f'(x).

when x > 0, f'(x) > 0.

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ANSWER: f(x) is increasing when f'(x) > 0, which happens on the intervals (-3,0), (0, oo).

—Note that f(x) is not increasing at x = 0, because f'(x)=0 there, which is not > 0

Oct 23, 2017

The interval of increasing of f(x) is x in (-3,0) uu (0,+oo)

Explanation:

We need

(uv)'=u'v+uv'

The function is

f(x)=x^3e^x

This is a product of two functions

u(x)=x^3, =>, u'(x)=3x^2

v(x)=e^x, =>, v'(x)=e^x

Therefore,

f'(x)=3x^2e^x+x^3e^x

=e^x(3x^2+x^3)

The critical points are when f'(x)=0

e^x(3x^2+x^3)=0

AA x in RR, e^x>0

3x^2+x^3=0, =>, x^2(3+x)=0

Therefore, the critical points are when

x=0 and x=-3

We can make a sign chart

color(white)(aaaa)Intervalcolor(white)(aaaa)(-oo,-3)color(white)(aaaa)(-3,0)color(white)(aaaa)(0,+oo)

color(white)(aaaa)f'(x)color(white)(aaaaaaaaaaa)-color(white)(aaaaaaaaaa)+color(white)(aaaaaaaa)+

color(white)(aaaa)f(x)color(white)(aaaaaaaaaaa)color(white)(aaaaaaaaaa)color(white)(aaaaaaaa)

The interval of increasing of f(x) is x in (-3,0) uu (0,+oo)

graph{x^3e^x [-12.66, 12.65, -6.33, 6.33]}