# Thread: max, min, inflections

1. ## max, min, inflections

. Let f(x) = x2ex
(a)Find all of the local maxima and minima
(b)find point of inflection
(c)determine largest interval in which f(x) is concave downward.
(d) evaluate lin x->-inf f(x) and lim x->inf f(X)

2. Originally Posted by gracy
. Let f(x) = x2ex
(a)Find all of the local maxima and minima
If a function is continous everywhere then the relative extrema occur either when it is not differenciable or the derivative is zero. The function is differenciable everywhere.
We search where it is zero.
$y=x^2e^x$
Product rule,
$y'=2xe^x+x^2e^x$
Now, $y'=0$ thus,
$2xe^x+x^2e^x=0$
Since, $e^x\not = 0$ divide by it,
$2x+x^2=0$
Thus,
$x(2+x)=0$
Thus,
$x=-2,0$.

That means the function is monotone on,
$x<-2,-20$
Check the signs of these intervals,
$x<-2 \to f'(x)>0 \to \mbox{increasing}$
$-2
$00 \to \mbox{increasing}$
By the first derivative test $x=-2$ is relative maximum.
By the first derivative test $x=0$ is relative minimum.

3. Hello, Gracy!

Let $f(x) \,= \,x^2e^x$

(b) Find point(s) of inflection
(c) Determine largest interval in which f(x) is concave downward.
(d) Evaluate $\lim_{x\to-\infty}f(x)$ and $\lim_{x\to\infty}f(x)$

(b) Inflection points occur where $f''(x) = 0.$
The second derivative is: . $f''(x)\:=\:e^x(x^2+4x+2)$
Since $e^x \neq 0$, we have: . $x^2 + 4x + 2\:=\:0$

Quadratic Formula: . $x\:=\:\frac{-4 \pm\sqrt{8}}{2}\:=\:-2 \pm\sqrt{2}$

Inflection points at: $x = -2 - \sqrt{2},\; -2 + \sqrt{2}$

(c) Testing values below, between and above the inflection points,
. . the graph is concave down on the interval: $(-2-\sqrt{2},\:-2 + \sqrt{2})$

(d) Evaluate limits

As $x$ approaches negative infinity, $x^2$ remains positive.
. . However, $e^x$ approaches 0.
Since $e^x$ has a greater "growth rate" than $x^2$,
. . the function approaches 0.
Therefore: . $\lim_{x\to-\infty}x^2e^x \;= \;0$

As $x$ approaches positive infinity, both $x^2$ and $e^x$ increase without bound.
Therefore: . $\lim_{x\to\infty}f(x)\;=\;\infty$