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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.

2 Answers 2

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  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 $2e^{3x}-1=0$.

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    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?2011-10-24
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    @Jordan: You *do* know, you just aren't thinking it through. You need to use logarithms.2011-10-24
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    Is it x=1/4? I rasied everything by ln.2011-10-24
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    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 $\ln(1/2)$, $\ln(1/2) = \ln(1)-\ln(2) = -\ln(2)$.2011-10-24
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    So it should be something like -.2301?2011-10-24
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    @Jordan: Don't approximate, give *exact answer*. The answer will be expressed in terms of $\ln(2)$. Every time you approximate, you introduce an error. Further computations using the approximation are approximations on the approximation, introducing further and further errors. The only value of $x$ for which $f'(x)=0$ is *exactly given* by $x=-\frac{\ln 2}{3}$.2011-10-24
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    Thank you, I can get it from here I understand the calculus for the most part I just keep forgetting all the other stuff.2011-10-24
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    @Jordan: I suspect you'll run into trouble on point 7, since you will need to find $f''$, you will need to solve $f''(x)=0$.2011-10-24
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    I am not too sure about the properties of e but I think the second degree derivative is $(6e^3x)/e^x + (2e^3x-1)/e^x$2011-10-24
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    @Jordan: No, that's wrong. Don't use the final formula of $f'$ to compute $f''$; I did that so that solving $f'(x)=0$ would be easier. Use $f'(x) = 2e^{2x} - e^{-x}$ instead.2011-10-24
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    Does that change the answer I get or does it just make it harder if I use the factored/reduced formula? I think I get $4e^{2x} - e^{-x}$2011-10-24
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    @Jordan: Doing the derivative *correctly* will give you the same answer, whether you do it from the first formula, or the last formula. But given your lack of facility with algebra, you want to minimize the opportunity for error. Your first attempt was incorrect; this attempt is *almost* right, but you've got a wrong sign.2011-10-24
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    Is the e supposed to be positive? I am not sure what I did wrong.2011-10-24
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    @Jordan: $(2e^{2x} - e^{-x})' = 2(e^{2x})' - (e^{-x})' = 2(2e^{2x})) - (-e^{-x}) = 4e^{2x} + e^{-x}$.2011-10-24
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    Is the derivative of $e^-x$ the same as $y=e^u$ and $u=-1(x)$ so then I get $-1(e^{-x})$?2011-10-24
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    @Jordan: The derivative of $e^{-x}$ *now* is exactly the same as the derivative of $e^{-x}$ when I did the computation in the second through fifth line of the display at the end of my post. So it is *still* equal to $-e^{-x}$, yes.2011-10-24
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    @Jordan just a note on why it's easier to use $f'(x)=2e^{2x}−e^{−x}$, it's because in general taking the derivative of a sum of functions is "easier" than taking the derivative of a product of functions. $\frac{1}{e^x}(2^{e3x}−1)$ is the product of $\frac{1}{e^x}$ and $2^{e3x}−1$, so you'll need to do something like use the product rule which is messier and provides you with more opportunity to make algebraic mistakes. $2e^{2x}−e^{−x}$ on the other hand is a sum, which means $( 2e^{2x}−e^{−x})'=(2e^{2x})'+(−e^{−x})'$, which is much easier to differentiate by hand.2012-12-09
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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?

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    According to wolfram that derivative is wrong.2011-10-24
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    @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.2011-10-24
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    How does the derivative turn into a $e^{3x}$? I dont follow at all.2011-10-24
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    @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.2011-10-24
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    The derivative is $e^{2x} + e^-1$ isn't it?2011-10-24
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    No. It is $2e^{2x}-e^{-x}$ which is the same as $e^{-x}(2e^{3x}-1)$ by the laws of indices, as given by Walfram. I do not understand what you dont get again.2011-10-24
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    I don't know what an indice is.2011-10-24
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/1639/discussion-between-smanoos-and-jordan)2011-10-24