algerbraically determine all critical values: F(x): (sq rt of (x-1))/ e^x?

Answer 1

Restating your question (for syntax purposes):

F(x) = sqrt(x-1) / e^x

The critical values are defined to be when F'(x) = 0 or when F'(x) is undefined. Let's solve for F'(x).

To solve for F'(x) we need to use the quotient rule.

F'(x) = [1/(2sqrt(x-1))*e^x - sqrt(x-1)*e^x]/(e^x)^2
F'(x) = (e^x - 2*(x-1)*e^x)/[(e^2x)*(2*sqrt(x-1))]
F'(x) = 1/2 * [1 - 2(x-1)]/(e^x * sqrt(x-1))
F'(x) = 1/2 * [2 - 2x]/(e^x * sqrt(x-1))

We then calculate when F'(x) = 0 or when F'(x) is undefined.
0 = 1/2 * [2 - 2x]/(e^x * sqrt(x-1))
0 = [2 - 2x]/(e^x * sqrt(x-1))

So
2 - 2x = 0
e^x * sqrt(x-1) = 0

2x = 2
sqrt(x-1) = 0

x = 1
x-1 = 0

Which yields x = 1 as a critical value.

Answer 2

Critical values are when the derivative = 0 or undefined.

F'(x) = ( e^x(1/2)(x-1)^(-1/2) + (x-1)^(1/2)e^x ) / e^(2x)

F' is undefined when the denominator, e^2x = 0. The graph of e ^x never equals 0, so it is never undefined.

F' = O when the numerator, e^x ( 1/2(x-1)^(1/2) + (x-1)^(1/2) ) = 0. Solve this for what x equals and you have the critical values.