# Thread: exact local max

1. ## exact local max

Let f(x)=x^80/e^x= x^80*e^-x. A graph of y= f(x) is shown below. There is clearly a local maximum point somewhere around x= 75 (actually the extree point is to the right of x=75). Also there appears to be at least one local minimum between x=-15 and x=45.

Use calculus to find the exact location of the local maximum that is near x=75 show that there is a local minimum at x=0.

I cannot show the grow however i can tell you that the y range goes by .5*10^116. However after that the exponents remain at 117 rather than 116. The x range goes by 15 starting at -30 going up to 150. Also I can see that the local max according to the graph is at coordinates (80, 3.1*10^117).

How would i figure this out? Would I find the slope of the tangent line?

2. Originally Posted by asweet1
Let f(x)=x^80/e^x= x^80*e^-x. A graph of y= f(x) is shown below. There is clearly a local maximum point somewhere around x= 75 (actually the extree point is to the right of x=75). Also there appears to be at least one local minimum between x=-15 and x=45.

Use calculus to find the exact location of the local maximum that is near x=75 show that there is a local minimum at x=0.

I cannot show the grow however i can tell you that the y range goes by .5*10^116. However after that the exponents remain at 117 rather than 116. The x range goes by 15 starting at -30 going up to 150. Also I can see that the local max according to the graph is at coordinates (80, 3.1*10^117).

How would i figure this out? Would I find the slope of the tangent line?
$f(x)=x^{80}e^{-x}$
$f'(x)=80x^{79}e^{-x}-x^{80}e^{-x}=x^{79}e^{-x}(80-x)$

Set $f'(x)=0$ and solve for $x$. You can use the second derivative test to see if it's a max or min:

$f''(x_0)>0\implies\min$
$f''(x_0)<0\implies\max$

3. so would i plug the derivative into my calculator and try to find the local max

4. Originally Posted by asweet1
so would i plug the derivative into my calculator and try to find the local max
You don't need a calculator to find the extrema. A calculator might be helpful to see whether they're local mins or maxes, but that's it.

$x^{79}e^{-x}(80-x)=0\implies x^{79}=0~\text{or}~80-x=0\implies x=0~\text{or}~x=80$