I am attempting to study for a test, but I forgot how to do all the stuff from earlier in the chapter.

I am attempting to find the intervals where $f$ is increasing and decreasing, min and max values and intervals of concavity and inflection points.

I have $f(x)= e^{2x} + e^{-x}$. I don't really know what to do. I know that I have to find the derivative, set to zero to find the critical numbers then test points then find the second derivative and do the same to find the concavity but I don't know how to get the derivative of this. Wolfram Alpha gives me a completely different number than what I get.

  1. Find the derivative of $f$.

  2. Find all the points where $f'$ does not exist or is equal to $0$. These are the only points where $f'$ can "change signs".

  3. Using the points from 2, determining on what intervals $f'\gt 0$ (that's where $f$ is increasing), and on what intervals $f'\lt 0$ (that's where $f$ is decreasing).

  4. Local maxima of $f$ occur at point where $f$ is defined, and $f'$ switches from positive (tangent lines that look like /) to negative (tangent lines that look like \).

  5. Local minima of $f$ occur at points where $f$ is defined, and $f'$ switches from negative (tangents \) to positive (tangents /).

  6. Find the second derivative of $f$.

  7. Repeat points 2 and 3 for $f''$. Intervals where $f''\gt 0$ are where $f$ is concave up; intervals where $f''\lt 0$ are where $f$ is concave down. (If you can't remember which is which, draw a concave up "cup", $\cup$, and notice that as you move left to right along the cup, the slope of the tangents increases; that means $f'$ is increasing, that means $f''\gt 0$; draw the same with a concave down "dome", $\cap$, to notice the slopes of the tangents decrease as we move left to right along the dome).

  8. Points of inflection are points where $f''$ changes from positive to negative, or from negative to positive.

For this particular function, you need to use the Chain Rule twice (well, you need to remember how to do derivatives, most of all).

$$\begin{align*} f'(x) &= \Bigl( e^{2x} + e^{-x}\Bigr)\\ &= \Bigl(e^{2x}\Bigr)' + \Bigl(e^{-x}\Bigr)'\\ &=e^{2x}(2x)' + e^{-x}(-x)'\\ &=e^{2x}(2) + e^{-x}(-1)\\ &=2e^{2x} - e^{-x}\\ &= 2e^{2x} - \frac{1}{e^x}\\ &= \frac{2e^{2x}e^x}{e^x} - \frac{1}{e^x}\\ &= \frac{2e^{3x} - 1}{e^x}\\ &= \frac{1}{e^x}(2e^{3x}-1). \end{align*}$$

This is always defined; and in order to be equal to $0$, you need $2e^{3x}-1=0$.

  • $\begingroup$ I get to $e^{3x} = 1/2$ and then I do not know what to factor any further. Can I raise everything to the -3x? $\endgroup$ – toby yeats Oct 24 '11 at 2:14
  • $\begingroup$ @Jordan: You do know, you just aren't thinking it through. You need to use logarithms. $\endgroup$ – Arturo Magidin Oct 24 '11 at 2:23
  • $\begingroup$ Is it x=1/4? I rasied everything by ln. $\endgroup$ – toby yeats Oct 24 '11 at 2:25
  • $\begingroup$ You don't "raise" things by "ln". You apply the logarithm to both sides. Again, this is algebra, not calculus. If $e^{3x}=1/2$, then $\ln(e^{3x}) = \ln(1/2)$. What is $\ln(e^{3x})$? As for $\ln(1/2)$, $\ln(1/2) = \ln(1)-\ln(2) = -\ln(2)$. $\endgroup$ – Arturo Magidin Oct 24 '11 at 2:28
  • $\begingroup$ So it should be something like -.2301? $\endgroup$ – toby yeats Oct 24 '11 at 2:29

This is how to find the derivative. If you are familiar with the derivative of the exponential function then notice that $\frac {df}{dx}=2e^{2x}-e^{-x}$. Set this zero to find the critical points. Thus, $2e^{2x}-e^{-x}=0$ which implies, $2e^{3x}-1=0$ (Multiply through by $e^{x}$), from which we get $e^{3x}=\frac {1}{2}$. So $x=\frac {1}{3}\ln(\frac{1}{2})=-\frac {1}{3}\ln2$. Now you can continue, right?

  • $\begingroup$ According to wolfram that derivative is wrong. $\endgroup$ – toby yeats Oct 24 '11 at 1:29
  • $\begingroup$ @Jordan. I just checked Walfram and it is the same thing. It only factorizes $e^{-x}$ out of the expression. Take a look at it well. Infact Walfram even gives the same answer for x as i have above. $\endgroup$ – smanoos Oct 24 '11 at 1:33
  • $\begingroup$ How does the derivative turn into a $e^{3x}$? I dont follow at all. $\endgroup$ – toby yeats Oct 24 '11 at 1:44
  • $\begingroup$ @Jordan. Just multiply through the equation by $e^{x}$ to get that. It does not change the equation because the right hand side is zero. $\endgroup$ – smanoos Oct 24 '11 at 1:47
  • $\begingroup$ The derivative is $e^{2x} + e^-1$ isn't it? $\endgroup$ – toby yeats Oct 24 '11 at 1:52

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