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When is the function defined by $f(x)=x^2+e^{-2x}$ increasing? I know you have to take the derivative and use certain values of $x$, but I am confused on how to do this particular problem, and I would love it if any could help. Any help is appreciated thanks.

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A differentiable function (such as this one) is increasing when its derivative is positive (and decreasing when its derivative is negative). Can you figure out when the derivative of $f$ is positive? –  Goos Apr 30 '13 at 22:53
    
is the derivative 2x + e^-2x? –  Captn Buzz Apr 30 '13 at 22:57
    
No it's not, see answer below. The minus two is there because we are taking the derivative of a composed function. –  justt Apr 30 '13 at 23:02

2 Answers 2

First take the derivative:

$f'(x)=2x-2e^{-2x}$

Now consider the values when the derivative is equal to 0:

$2x-2e^{-2x}=0$

Now the function is increasing whenever $f'(x)$ is positive, and decreasing when it is negative. I.e., $f$ is increasing when: \begin{align*} 2x &> 2e^{-2x}\\ x &> e^{-2x}\\ \ln(x) &> -2x \end{align*}

As @CommonerG says below, you can compute a decimal approximation to this root.

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$\frac{\delta f(x)}{\delta x}=-2e^{-2x}+2x$

Thus, the function is increasing when:

$2x>2e^{-2x}$

$x>\frac{1}{e^{2x}}$

A computational approach to solving this inequality provides the condition:

$x>.426$

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