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I’m in an IB Math class and we are working on some calculus problems but I wanted to get extra practice so this is a problem in my book. The number in parenthesis next to the parts are the “marks” we get for the question if we get it right. So, usually that’s about how much work we have to show or how many steps to take to complete the problem.

The following diagram shows the graph of $f(x) = e^{-x^2}.$ The points $A, B, C, D,$ and $E$ lie on the graph. Two of these are points of inflection.

diagram here

a) Identify the two points of inflection. (2)

Isn’t this $B$ and $D$? Since that’s where the function is changing signs?

b) Find $f’(x)$. (2)
From the comments, I was told this was the derivative: $-2xe{x^2}$

c) Show that $f’’(x) =(4x^2 - 2)e^{-x^2}$. (2)

This is confusing me a bit. Was the Product Rule used here?

d) Find the $x$-coordinate of each point of inflection. (4)
Do we set the second derivative to 0 and solve from there?

$0 = (4x^2 - 2)e^{-x^2}$

$0 = 4x^2 -2$

$2 = 4x^2$

$1/2 = x^2$

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    $\begingroup$ The derivative is not $2e^{-x}$. This is not just a bit wrong! So try it again. $\endgroup$ – TonyK Dec 13 '18 at 21:32
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    $\begingroup$ For $a$, yes it's B&D. But you have to say "what" is changing. The thing that is changing is the concavity of the graph, or in other words the sign of the second derivative. For $b$, the derivative is actually $-2xe^{x^2}$. $\endgroup$ – Ovi Dec 13 '18 at 21:33
  • $\begingroup$ For the inflection points use second derivative, for the derivative recall how a derivative of a composite function is obtained $\endgroup$ – roman Dec 13 '18 at 21:33
  • $\begingroup$ @Ella No, you can see that $e^{-x^2}>0\ \forall x\in\Bbb R$, so the function doesn't change signs. It is the second derivative of the function that is changing signs at $B, D$. $\endgroup$ – Shubham Johri Dec 13 '18 at 22:14
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Isn’t this $B$ and $D$? Since that’s when it’s changing?

Yes, $B$ and $D$ are the points of inflection. For the sake of clarity, the "it" you're talking about is the concavity of the graph of the function, as rightly pointed out in one of the comments. These are the points where the function goes from being convex to concave and vice-versa.

As for the second part, note that the power rule states $\frac{d}{dx}(x^a)=ax^{a-1}$, where $a$ is a constant. In the case of $e^{-x^2}$, the exponent $-x^2$ is variable. You will have to use the chain rule here:

$\displaystyle\frac d{dx}[h(g(x))]=\frac d{d[g(x)]}[h(g(x))]\times\frac d{dx}[g(x)] =h'(g(x))\times g'(x)$

It simply means that when you differentiate the composition of two functions, you first have to differentiate the outer function with respect to the inner function and multiply that with the derivative of the inner function.

In your case, take $h(x)=e^x, g(x)=-x^2\implies h(g(x))=h(-x^2)=e^{-x^2}$

$\displaystyle\frac d{dx}(e^{-x^2})=\frac d{d(-x^2)}(e^{-x^2})\times\frac {d(-x^2)}{dx}=\frac d{dm}(e^{m})\Big|_{m=-x^2}\times(-2x)=-2xe^m=-2xe^{-x^2}$

Note that you could have also taken $h(x), g(x)$ to be $e^{-x}, x^2$ but the final answer will be the same.

Now for calculating the second derivative of $e^{-x^2}$, you will have to differentiate the first derivative $-2xe^{-x^2}=a(x)\times h(g(x))$ using the product rule.

$f''(x)=a'(x)\times h(g(x))+a(x)\times[h(g(x))]'$, where $a(x)=-2x, h(g(x))=e^{-x^2}$.

You will notice that we have already calculated the term $[h(g(x))]'=f'(x)$ above.

As for part $(d)$, you are on the right track.

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  • $\begingroup$ Thank you! For part d, it asks for the coordinate of each point of inflection. I got $x$ = the square root of $1/2$. Should I use the decimals for that? Since there are 2, it would be the positive and negative value, right? $\endgroup$ – Ella Dec 13 '18 at 22:50
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    $\begingroup$ Yes, $x=\pm\frac1{\sqrt2}$. Remember, co-ordinate means both the $x$ and $y$ values! Don't forget to find $f(\pm\frac1{\sqrt2})$. I don't think you need to use decimals, radicals are preferred in calculus $\endgroup$ – Shubham Johri Dec 13 '18 at 22:52
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Some help:

a) Yup, that seems right!

b) Here you should use the chain rule and the power rule. Think of it as a composition of two functions. Perhaps as $g(x)=-x^2$ and $h(x)=e^x$ will make more sense. Your function will then be $f(x)=h(g(x)).$

c) Once you have the correct result for $f'(x),$ you just can use the product rule to compute $f''(x).$

d) The first derivative tells you how fast your function is increasing or decreasing. Going from $A$ to $B$ you see the rate at which it increases goes up. Then at $C$ it does not grow at all and if you go further it starts to sink again. It sinks faster and faster until we reach $D$. From then on it still sinks but the rate at which it does so goes to zero. - So for some $x,$ the derivative $f'(x)$ is positive and for others negative. In the positive part there is a point where it is at a maximum and for the negative part there is a point where it is at a minimum. At both point we have that the change in the first derivative with respect to $x$ is zero. Those are your points of inflection. This is nothing but finding out for what $x_0$ your second derivative is zero. I.e. $f''(x_0)=0.$

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You are correct, the points of inflection are B and D.

In order to take the derivative of this function, we must utilize multiple rules. First and foremost, this is an exponential function. It appears by your attempt to use the power rule that you are unfamiliar with a few derivative rules. Lets take a look at this.

The derivative of a function is defined by the following: $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ Let us forget about the argument of the exponential, and call it all $x$ $$\lim_{h\to 0}\frac{e^{x+h}-e^x}{h}$$ $$\lim_{h\to 0}\frac{e^{x}e^h-e^x}{h}$$ $$\lim_{h\to 0}\frac{e^{x}(e^h-1)}{h}$$ $$e^x\lim_{h\to 0}\frac{e^{h}-1}{h}$$ Looking at the graph of this function, it is evident that the limit evaluates to 1, implying that $\frac{d}{dx}e^x=e^x$

Since the argument of the exponential is not simply $x$, we must use the chain rule as well. I won't go into much detail on this one, but the chain rule states that $$\frac{d}{dx}f(g(x))=f'(g(x))*g'(x)$$ Since we can think of your function like this, $e^{f(x)}$, where $f(x)=-x^2$, and using our newfound derivative of $e^x$, we can say

$$\frac{d}{dx}e^{-x^2}=e^{-x^2}*(\frac{d}{dx}(-x^2))$$

and since we know the derivative of $-x^2$ to be $-2x$, we can conclude that $$f'(x)=-2xe^{-x^2}$$

From here, I will leave it to you to apply the product rule (with which you have demonstrated familiarity) along with our old friend, the chain rule, to verify part (c).

You did part (d) correctly.

Good luck with it, and really try to take in the beauty of calculus! It truly is breathtaking, and I am reminded of this every time I get the opportunity to answer a calculus question.

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