Science:Math Exam Resources/Courses/MATH101/April 2005/Question 08 (b)

MATH101 April 2005

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Question 08 (b)

An unknown continuous function $f(x)$ satisfies $f(0)=0$, $f(4)=8$, and $2x\leq f(x)\leq 6x-x^{2}$ for $0\leq x\leq 4.$ Also, $f(x)$ is nondecreasing on this interval, i.e. it satisfies $f(c)\leq f(d)$ for all real numbers $c$ and $d$ with $0\leq c\leq d\leq 4.$ Let $I$ be the value of definite integral $\int _{0}^{4}f(x)\,dx.$ (b) Find the smallest and largest posible values for $I$. Give explicit functions $f(x)$, satisfying the conditions above, that yield these smallest and largest values.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Try actually making $f(x)=2x$ and $f(x)=6x-x^{2}$. What problems does doing this give (if any)? How can we overcome them?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The minimum is given by $g(x)=2x$ and the maximum is given by $h(x)={\begin{cases}6x-x^{2}&\quad x\leq 2\\8&\quad 2\leq x\leq 4\end{cases}}$ To show that $g(x)$ is the minimum, we note that our $f(x)$ is a nondecreasing function, $f(0)=0$ and we also require that $2x\leq f(x)$. This last condition means that the absolute smallest $f(x)$ can be is $f(x)=2x$ and it turns out that this function also has the first two desired properties so this function works and is minimal. For the second function, we note that our $f(x)$ is a nondecreasing function; we also have $f(4)=8$ and lastly that $f(x)\leq 6x-x^{2}$. Again to make the function maximal and nondecreasing, we really would like $f(x)=8$ however this breaks the upper bound condition. So we pick the function $6x-x^{2}$ until we reach the value of 8 and then pick the function $f(x)=8$ until the end where we also need to hope that $8\leq 6x-x^{2}$ on this interval (so that we are not breaking the upper bound condition. Let $y=6x-x^{2}$. Notice that $y'=6-2x>0$ when $x<3$. This means that the function is non-decreasing until $x=3$. Furthermore, notice that $8=6x-x^{2}$ when $0=x^{2}-6x+8=(x-2)(x-4)$ and so this occurs when $x=2$ and $x=4$. Notice that since this is a parabola, it has only one critical point (which we computed to be at $x=3$ and so indeed between $2\leq x\leq 4$, we have that $8\leq 6x-x^{2}$. Thus, the function $h(x)$ chosen above has the desired properties we require and thus gives the maximum value for I. Computing these values $\int _{0}^{4}g(x)\,dx=\int _{0}^{4}2x\,dx=x^{2}|_{0}^{4}=16$ and $\int _{0}^{4}h(x)\,dx=\int _{0}^{2}6x-x^{2}\,dx+\int _{2}^{4}8\,dx=\left.\left(3x^{2}-{\frac {x^{3}}{3}}\right)\right|_{0}^{2}+16=12-{\frac {8}{3}}+16={\frac {76}{3}}$ completing the question.

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