# 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?) Aug 15, 2019 at 17:01

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 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 $g(x)?$ Aug 15, 2019 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! Aug 15, 2019 at 17:29
• So for the interval at which is overlaps, how would I treat that case? Aug 15, 2019 at 17:36
• @Aurora Borealis Well you would just split up the integral appropriately. For example, $f(2)$=$\int_{-4}^{0}$($-1$)$dt$+ $\int_{0}^{2}$($t-1$)$dt$ Aug 15, 2019 at 17:41
• Oh ahah yes, thank you ! Aug 15, 2019 at 17:41

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