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I need to find the extrema of $f(x)=(x^2+2x-7)e^{|x+2|}$. The first derivative is $e^{|x+2|}(2x+2+(x^2+2x-7)(|x+2|)')$. Looking at two cases where $x<-2$ or $x>-2$ gives two extrema - at the points -3 and 1. And now comes the real question: Is there an extremum at the point -2 and why? If we replace x with -2, the first derivative equals -2. But the fact that the derivative there isn't $0$ doesn't mean there isn't an extremum. In fact, if I take a look at the graphic of the function, it is obvious that in the region of -2, the value of the function for $x=-2$ is the highest. So, according to the definition, there should be a local maximum. Which one is true? Is $x=-2$ a local maximum and why?

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  • $\begingroup$ How did you get extrema "at the points $=3$ and $1$"? [show what you did] $\endgroup$
    – coffeemath
    Commented Dec 24, 2020 at 17:08
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    $\begingroup$ Case1: $x<-2$, therefore the module brackets are removed and the content is taken with a negative sign. (instead of $|x+2|$, we get $-(x+2)=-x-2. (-x-2)'=-1$, etc. Case2: $x>-2$, therefore the module brackets are removed and the content is taken with a positive sign. (instead of $|x+2|$, we get $(x+2). (x+2)'=1$, etc. $\endgroup$ Commented Dec 24, 2020 at 17:13

2 Answers 2

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The derivative of $|x+2|$ is $\dfrac {x+2}{|x+2|} = sgn(x+2) := \begin{cases} 1 &\text{ for } x+2 > 0\\-1 &\text{ for } x+2 < 0 \end{cases}$ and is not defined for $x=-2$. Therefore the derivative of $f(x)$ is also not defined for $x=-2$. (Some textbooks include these points where $f'(x)$ is not defined as critical points; regardless one must check whether an extremum exists at these points.)

The first derivative test can still be applied by taking left/right limits. Note that:

$$\lim_{x \to -2^-} f'(x) = \lim_{x \to -2^-} e^{|x+2|}\left(2x+2+(x^2+2x-7)\dfrac {x+2}{|x+2|}\right)=e^0(-4+2+(4-4-7)(-1))=5$$

$$\lim_{x \to -2^+} f'(x) = \lim_{x \to -2^+} e^{|x+2|}\left(2x+2+(x^2+2x-7)\dfrac {x+2}{|x+2|}\right)=e^0(-4+2+(4-4-7)(1))=-5$$

This shows that the function is increasing before $-2$ and decreasing after $-2$, and thus there is a local maximum at $x=-2$.

A similar analysis can be carried out on a simpler function: $|x|$. It is not differentiable at $x=0$, but has a local minimum there.

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Hint: You can simplify a lot the problem taking the $\log$ of your function and studying the extrema of the new function.

Indeed $f(x)= (x^2 +2x -7)e^{|x+2|}$ has the same extrema of $g(x)= \log[(x^2 +2x -7)e^{|x+2|}]$ and you can easily see that $$g(x)= \log[(x^2 +2x -7)e^{|x+2|}] = \log(x^2 +2x -7) + |x+2|$$

The extrema are the same since the log is a strictly increasing function and the composition of any function with a strictly increasing function does not change the extrema.

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