find the absolute max and min of f(x)= e^x/x in [1/2,3]
- Thread starter Ashley5
- Start date
Re: find the absolute max and min
Hello, Ashley!
I guess you need an Algebra Review . . .
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: \(\displaystyle \:\begin{array}{cc}\text{The absolute maximum is: }& \left(3,\:\frac{e^3}{3}\right) \\ \\ \\
\text{The absolute minimum is: } & \left(1,\:e\right) \end{array}\)
Hello, Ashley!
I guess you need an Algebra Review . . .
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: \(\displaystyle \:\begin{array}{cc}\text{The absolute maximum is: }& \left(3,\:\frac{e^3}{3}\right) \\ \\ \\
\text{The absolute minimum is: } & \left(1,\:e\right) \end{array}\)
