Criteria for the maximum of $f(x) = axe^{bx}$ at $(2,1)$

Question

I have this question and have been trying for a long time to fully solve it, hopefully you guys can help me. Here is the question:

Consider the function $f(x) = axe^{bx}$, where $a$ and $b$ are constants and real numbers. Find the possible value(s) of $a$ and $b$ such that $f(x)$ has an absolute maximum of $(2, 1)$ on the interval $[0, 50]$. Give your answer in exact form.

I got one answer, but can't seem to prove or find out other possible answers. Here is my solution for one possible answer:

$f(x) = axe^{bx}$

• $1=f(2) =2ae^{2b}$

  $ae^{2b}=\frac12$

• $f'(x) = ae^{bx}(1+xb)$

  $0=f'(2) = ae^{2b}(1+2b)=\frac12(1+2b)$

  $b = -\frac12$

$ae^{2(-1/2)}=\frac12$

$a =\frac e2$.

So one of my possible answers is $a=\frac e2$ and $b=-\frac12$. Now my only problem is the question asked for the possible value(s), so I don't know if there is more than one possible answer. Then question did specify on the interval from $[0,50]$. So I don't know if there is a function that is greater in the negative, but has a maximum on the interval $[0,50]$ on the positive end, or sort of a U-shaped function where it has a maximum at $(2,1)$ goes below, but then possible goes back up after $x=50$, which would still satisfy the criteria.

Just looking for some help if I got the only answer, or if there are more answers.

*Note: the question did say that $a$ and $b$ are constants and real numbers.

Answer 1

Questions that say something along the lines of "possible value(s)" actually mean: "find the one possible value of $a$ and $b$ or if multiple values work list all of them". This is worded like that so that the question doesn't hint on how many answers you should get, it's vague on purpose!

The work you provided is indeed correct and indeed, there is only one possible combination of values for $a$ and $b$, namely as you mentioned: \begin{align} a= \dfrac{e}{2} && \text{and}&& b=-\dfrac{1}{2} \end{align}

Answer 2

Your own calculation $$f'(x)=ae^{bx}(1+xb)$$ shows that $f$ has a unique critical point (except of course if $a=0$ or $b=0$, in which case $f$ is constant). So, the answer to your question about the possibility of more complicated shapes for the graph of $f$ is: no.

Moreover, this critical point is an absolute extremum on $\Bbb R$ and is at $$\left(-\frac1b,-\frac abe^{-1}\right)=(2,1)$$ iff $b=-\frac12$ and $a=-eb=\frac e2$, as you found. (It is then de facto also an absolute extremum on any interval containing $2$.)

The only thing you forgot to check, to make sure that this unique possible solution is indeed a solution (i.e. that there is one solution and not zero), is that this extremum is a maximum. You can check more generally that if $ab<0$ then the extremum of $f$ at $x=-1/b$ is a maximum, by looking at the sign of $f'(x)=abe^{bx}(x+1/b)$ when $x<-1/b$ and when $x>-1/b$.

Answer 3

By differentiation (ProductRule, simplification) it is found $$(x_m,y_m)=(-1/b,-a/(be))$$ when equated to $(2,1)$ we obtain $$ (a,b)= (e/2, -\frac12)$$ Second derivative sign test shows max at this choice.

Your work is correct and it appears to me that there need not be any doubt . If a second value is admissible, it will surely come out here as a double or multiple root.

Confirmed by function graph also: