Question

F:R—>R,f(x)=x²eˣ,Determine intervals in which the given function are strictly increasing or strictly decreasing.

Answer 1

Given, $\bf{f:\mathbb{R}\to\mathbb{R}}$, $\bf{f(x)=x^2e^x}$

differentiate with respect to x,

$f'(x)=x^2.\frac{de^x}{dx}+e^x\frac{dx^2}{dx}$

$f'(x)=x^2.e^x+e^x.2x$

$f'(x)=x.e^x(x+2)$

now, f'(x) = 0

$x.e^x(x+2)=0$

x = 0 and -2

case 1 :- x < -2 , f'(x) > 0
so, function is strictly increasing in (-∞, 2)

case 2 :- -2 < x < 0 , f'(x) < 0
so, function is strictly decreasing in (-2,0)

case 3 :- x > 0 , f'(x) > 0
so, function is strictly increasing in (0, ∞)

finally , function is strictly increasing in (-∞, 2) U (0, ∞) while function is strictly decreasing in (-2,0)

Answer 2

Dear student:

Given: R—>R,

f(x)=x²eˣ

For determining the intervals in which f is increasing and decreasing.

Find derivative of f(x)

Then see the derivative in which f is positive and negative.

If it is positive then f is increasing

And if it is negative then f is decreasing.

See the attachment.