Derivative of f(x) = e^(x^2 - 2x), critical numbers, etc

[Jade]

I need to find the derivative of f(x) = e^(x^2 - 2x), then find the critical numbers, and then find the maximum and minimum on the intervals [-1, 2] and [-3, 0]

I believe the derivative would be f'(x) = e^(x^2 - 2x) (2x - 2)

I believe the critical numbers would be x = 1, 2

Then to find the maximum and minimums you would plug in the test numbers into the derivative correct???

 

[soroban]

Hello, Jade!

I need to find the derivative of \(\displaystyle f(x) \:= \:e^{x^2 - 2x}\) and find the critical numbers.
Then find the maximum and minimum on the intervals [-1, 2] and [-3, 0]

I believe the derivative would be: \(\displaystyle f'(x) \:= \:e^{x^2 - 2x}(2x\,-\,2)\) . . . right!

I believe the critical numbers would be: \(\displaystyle x \:= \:1,\,2\) . . . no

Then to find the maximum and minimums
you would plug in the test numbers into the derivative, correct?
No, pluggging into the derivative would give you 0.

Since \(\displaystyle e^{^{x^2-2x}}\) can never equal 0, the only critical value is \(\displaystyle x\,=\,1\)


On the interval \(\displaystyle [-1\,2]\), test the endpoints: \(\displaystyle \,f(-1)\:=\:e^3,\;f(2)\:=\:1\)

Since \(\displaystyle f(1)\,=\,\frac{1}{e}\), the maximum is \(\displaystyle \L e^3\), the minimum is \(\displaystyle \L\frac{1}{e}\)


On the interval \(\displaystyle [-3,\,0]\), test the endpoints: \(\displaystyle \,f(-3)\:=\:e^{15},\;f(0)\:=\:1\)

The maximum is \(\displaystyle \L e^{15}\), the minimum is \(\displaystyle \L1.\)

 

[Jade]

Thank you