Differentiate Exponents

• May 30th 2007, 12:05 PM
confusedagain
Differentiate Exponents
Find first and second derivatives, critical number, inflection points.

f(x) = xe^x
f'(x) = ('x) (e^x) + x(e^x)('x)
f'(x) = e^x+xe^x
f'(x) = e^x(1+x)
Critical number x = -1

f"(x) = e^x(1) + (1+x)(e^x)
f"(x) = e^x +e^x(1+x)
Critical Number x=-1

I hope I am close to the correct answer.
• May 30th 2007, 12:12 PM
Jhevon
Quote:

Originally Posted by confusedagain
f"(x) = e^x(1) + (1+x)(e^x)
f"(x) = e^x +e^x(1+x)
Critical Number x=-1

I hope I am close to the correct answer.

first of all, we do not find critical points from the second derivative, we use the second derivative to find concavity and possible inflection points. second of all,
$\displaystyle f''(-1) = \frac {1}{e} \neq 0$
• May 30th 2007, 01:42 PM
Soroban
Hello, confusedagain!

A small error . . .

Quote:

Find first and second derivatives, critical number, inflection points.
. . $\displaystyle f(x) \:= \:xe^x$

. . $\displaystyle f'(x) \:= \:1\cdot e^x + x\cdot e^x \:=\:e^x(1 + e^x)$

Then: .$\displaystyle e^x(1+x) \:=\:0\quad\Rightarrow\quad x = -1$

. . Critical point: $\displaystyle \left(-1,\:-\frac{1}{e}\right)$

$\displaystyle f''(x) \:= \:e^x + (1+x)e^x \:=\:e^x(1 + 1 + x) \:=\:e^x(2 + x)$

Then: .$\displaystyle e^x(2 + x) \:=\:0\quad\Rightarrow\quad x = -2$

. . Inflection point: .$\displaystyle \left(-2,\:-\frac{2}{e^2}\right)$

• May 30th 2007, 02:51 PM
Krizalid
Hey Soroban

If you prefer, use e^x(2 + x)=0\implies x=-2, this yields $\displaystyle e^x(2 + x)=0\implies x=-2$