Find the equation of a function whose graph has a relative minimum when x=2 and relative maximum when x=5.
This is the question and it's odd because usually we have to find the maximum and minimum when given a function.
Anyways, I'm not sure what to do as I haven't done a problem like this yet. I know that y'=0 when x=2, 5, and that y''>0 when x=2 and y''<0 when x=5. Any help is appreciated, really the starting point is needed, thanks in advance.
Best Answer
There are many such functions. If $f$ is a nice such function, then $f'(2)=0$ and $f'(5)=0$. We look for a function that works. Suppose that $f'(x)=(x-2)(x-5)$. Then $f'(x)$ is positive for $x\lt 2$, negative strictly between $2$ and $5$, and positive for $x\gt 5$.
So $f$ is increasing up to $2$, then decreasing up to $5$, then increasing: so we have a relative maximum at $x=2$, and a relative minimum at $x=5$. Not quite right!
The fix is easy. Let $f'(x)=-(x-2)(x-5)$. Then $f'(x)$ is negative for $x\lt 2$, positive strictly between $2$ and $5$, and then negative. So we have a relative minimum at $x=2$ and a relative maximum at $x=5$.
So $f'(x)=-(x-2)(x-5)=-x^2+7x-10$ will do the job. Can we think of a function whose derivative is this? Sure, $f(x)=-\frac{x^3}{3}+\frac{7x^2}{2}-10x$.
Remark: I deliberately avoided the second derivative test alluded to in the OP. This is because in my experience, first year calculus students who use the second derivative test tend to get into difficulties more than students who look at the behaviour of the first derivative.