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This is a problem I did in one of my test in high school, I completely forgot how to do this, there were two questions stated a) and b).

My attempt will be to try and find the function of the graph and integrate, then I would try to differentiate it and set it to 0, and try to find that value of of x? I am not quite sure how to fo part a), I think I can try to do part b) which is just integrate on that interval of the function.

Could anyone help me? Thank you

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  • $\begingroup$ The function $g(x)$ is differentiable, so to minimize it, try solving $g'(x)=0$ and checking solutions as well as endpoints of the interval on which it’s given. (Do you remember how to differentiate an integral with respect to the upper limit of integration?) $\endgroup$
    – Steve Kass
    Aug 15, 2019 at 17:01

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You can set $g'(x)=\dfrac{d}{dx}\int_{-4}^xf(t)dt$, which is $\dfrac{d}{dx}(F(x)-F(-4))$, if we let $F$ be an antiderivative of $f$. Then the derivative is clearly just $f(x)$.

So, where does $g'(x)=f(x)=0$? According to the graph, this occurs for $x=1$.

Note that part (b) is asking for a function that gives the area bounded by $f$ and the $x$-axis from $-4$ to $x$. You can easily plot points to see that, for example, $(0,-4)$ is on the graph.

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For part (a), we have that $g'(x) = f(x)$.

$g(x)$ has critical points whenever $g'(x)$= $f(x)$ =$0$. From the graph, this would correspond with $x=1$.

Since $g''(x)$=$f'(x)$, and $f'(1)$>$0$ (Why?), we know that $g''(1)$>$0$, so we know that $g(x)$ has a local minimum at $x=1$.

To test whether this is indeed the global minimum of $g(x)$, we must check the value of $g(x)$ at $x=1$ and also check the value of $g(x)$ at the endpoints of the interval.

$g(1)$=$\int_{-4}^{1}$ $f(t)$$dt$=$-4.5$

$g(-4)$=$\int_{-4}^{-4}$ $f(t)$$dt$=$0$

$g(4)$=$\int_{-4}^{4}$ $f(t)$$dt$=$0$

(Check these for yourself!).

So it is clear that $g(x)$ has a global minimum at $x=1$.

For part (b), there is no clear-cut solution. You might consider plotting critical points or in the worst-case scenario, you can just plot some arbitrary points.

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  • $\begingroup$ this was very helpful. I perfectly understand a). For b) can we assume that the function is$ f(x)=-1$ from $x\in [-4,0]$ and $f(x)=x-1$ from $x\in [0,4]$, and integrate these two functions separately and sketch it on those intervals for $g(x)?$ $\endgroup$ Aug 15, 2019 at 17:25
  • $\begingroup$ @Aurora Borealis I believe that this method would work. However, you just have to be careful because on the interval $x$ $\epsilon$ [$0$,$4$], you will be integrating both functions on their respective intervals. Anyways, I am glad to be of assistance! $\endgroup$
    – JG123
    Aug 15, 2019 at 17:29
  • $\begingroup$ So for the interval at which is overlaps, how would I treat that case? $\endgroup$ Aug 15, 2019 at 17:36
  • $\begingroup$ @Aurora Borealis Well you would just split up the integral appropriately. For example, $f(2)$=$\int_{-4}^{0}$($-1$)$dt$+ $\int_{0}^{2}$($t-1$)$dt$ $\endgroup$
    – JG123
    Aug 15, 2019 at 17:41
  • $\begingroup$ Oh ahah yes, thank you ! $\endgroup$ Aug 15, 2019 at 17:41
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For the first question, you have to remember the Fundamental Theorem of Calculus, which states:

$$F(x) = \int_a^xf(t)\text{ d}t$$ $$F'(x)=f(x)$$

Obviously the first integral fits this description, so we can say that $g'(x)=f(x)$. Now, to find the extrema of $g(x)$, we have to find a critical point. A critical point is a point where $g'(x)=0$, and we know that $f(x)=g'(x)$, so we can look at the graph of $f(x)$. The function $f(x)=0$ at $x=1$, and by the first derivative test (if a derivative goes from negative to positive at a critical point, that critical point it a minimum), we can tell that $x=1$ is the minimum of $g(x)$. Now, as was stated above, we have to evaluate the function at the endpoints of $f(x)$ to see if $x=1$ is a global minimum. Once you do this, you'll find that $x=1$ is indeed the global minimum of $g(x)$.

For the second question, I would suggest that you remember the rules of curve sketching:

  • When the first derivative is negative, the function is decreasing.
  • When the first derivative is positive, the function is increasing
  • Find the maxima of the function (we just did that)
  • Plot some points as a guide if necessary
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