# Solving a maximum minimum problem

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

• 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?) – Steve Kass Aug 15 '19 at 17:01

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.

• 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 $g(x)?$ – Aurora Borealis Aug 15 '19 at 17:25
• @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! – JG123 Aug 15 '19 at 17:29
• So for the interval at which is overlaps, how would I treat that case? – Aurora Borealis Aug 15 '19 at 17:36
• @Aurora Borealis Well you would just split up the integral appropriately. For example, $f(2)$=$\int_{-4}^{0}$($-1$)$dt$+ $\int_{0}^{2}$($t-1$)$dt$ – JG123 Aug 15 '19 at 17:41
• Oh ahah yes, thank you ! – Aurora Borealis Aug 15 '19 at 17:41

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.

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