Question

I’m trying to find the absolute Max and min, I can’t figure out how to do the derivative, thanks

Answer 1

Explanation:

Step 1: Take the First Derivative This means only differentiate once.

Disclaimer: Since absolute value only take positve outputs and quadratics only take positve outputs, we can get rid of the absolute value signs so we now have

`e {}^{ {x}^(2) - 1 }`

We have the function x^2-1 composed into the function e^x.

So we use chain rule

Which states the derivative of a function composed is the

derivative of the main function times the derivative of the inside function.

So the derivative of the main function is

`(d)/(dx) (e {}^(x) ) = e {}^(x)`

Then we replace x with x^2-1

`e {}^{ {x}^(2) - 1}`

Then we take the derivative of the second function which is 2x so qe multiply them

`e { }^{ {x}^(2) - 1 } 2x`

Step 2: Set the equation equal to zero.

`e {}^{x {}^(2) - 1} 2x = 0`

Since e doesn't reach zero. We can just set 2x=0.

`2x = 0 = x = 0`

So the critical point is 0.

Since e^x will never reach zero

Since 0 is the only critical point, this where the max or min will occur at.

Next we pick any numbergreater than zero, and plug them in the derivative function which gives us a positve number.

Any pick less than zero will give us a negative number.

Since the function is decreasing then increasing, we have a minimum.

Since 0 is the only critical point, we have a absolute minimum at 0.

To find the y coordinate, plug 0 in the orginal function.

Which gives us

`e {}^{ {0}^(2) - 1 } = e {}^( - 1) = (1)/(e)`

So the minimum occurs at

(0,1/e).