Local/Global Max and Min question

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njoyce

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Aug 1, 2006
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For f(x) = x^2 + (e^x)^2 and -0.5<= x >=3, find the values of x for which
A. f(x) has a local minimum.
B. f(x) has a local maximum.
C. f(x) has a global minimum.
D. f(x) has a global maximum.

List the values from smallest to largest, seperated by commas.

I'm sure I'm making this harder than it is, but e raised to anything confuses me. This is what I've done so far.

f'(x) = 2x + 2x(e^x)^2
now simplify to 2x[1 + (e^x)^2]

I know I have to solve for x, but the e raised to the x which is squared confuses me. I would really appreciate any help.

Thanks!!
 
First note that \(\displaystyle \L
\left( {e^x } \right)^2 = e^{2x}\).

Thus \(\displaystyle \L
\begin{array}{rcl}
f(x) & = & x^2 + e^{2x} \\
f'(x) & = & 2x + 2e^{2x} \\
\end{array}\)
 
Thank you so much for your prompt reply. I don't think I typed the function correctly when I look at your reply.

the e is raised to the x power and the x is squared. It's all one term. The e is raised to a power and that power is also raised to a power.


f(x) = x^2 + e^x^2
 
\(\displaystyle \L
\begin{array}{rcl}
f(x) & = & x^2 + e^{x^2 } \\
f'(x) & = & 2x + 2xe^{x^2 } \\
\end{array}.\)

Suggestion: GRAPH the function.
You will see the answer.
 
Thanks. I was able to graph the function. Can you help me work through the algebra to solve for x? I have to be able to show the work and I'm getting lost on the algebra.
 
The minimum occurs at x=0.
Check the second derivative.
 
Just factor it. 2*x*(stuff that can't be zero or negative)
 
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