find the absolute max and min of f(x)= e^x/x in [1/2,3]

[Ashley5]

find the absolute max and min of f(x)= e^x/x in [1/2,3]

I started taking the derivative but I don't know how to finish it. Anything will be helpful. Thank you!!

f'(x)= (x)e^x-e^x(1)/x^2

 

[soroban]

Re: find the absolute max and min

Hello, Ashley!

I guess you need an Algebra Review . . .

Find the absolute maximum and minimum of: \(\displaystyle \L\,f(x)\:=\:\frac{e^x}{x}\,\) on [½, 3]

The derivative is: \(\displaystyle \L\:y' \;=\;\frac{x\cdot e^x \,-\,e^x\cdot1}{x^2} \;=\;\frac{e^x(x\,-\,1)}{x^2}\)

If $\displaystyle y'\,=\,0$, we have: $\displaystyle \:x\,-\,1\:=\:0\;\;\Rightarrow\;\;x\,=\,1$

. . Then: $\displaystyle \,y \:=\:\frac{e^1}{1} \:\approx\:2.718$


Test the endpoints . . .

At $\displaystyle x\,=\,\frac{1}{2}$, we have: $\displaystyle \:y \:=\:\frac{e^{\frac{1}{2}}}{\frac{1}{2}} \:\approx\:3.297$

At $\displaystyle x\,=\,3$, we have: $\displaystyle \:y \:=\:\frac{e^3}{3} \:\approx\:6.695$

Therefore: \(\displaystyle \:\begin{array}{cc}\text{The absolute maximum is: }& \left(3,\:\frac{e^3}{3}\right) \\ \\ \\
\text{The absolute minimum is: } & \left(1,\:e\right) \end{array}\)

 

[Ashley5]

yea!! Thank you!! I get it now!