Hi Anjela,
As you know, the critical values of a function are the x-values that make the first derivative equal to 0 or that make the first derivative undefined.
Given the equation:
f(x) = 15 x^17/3 - 85 x^(1/2),
the first derivative (using the Power Rule on both terms) is:
f '(x) = 15*(17/3) x^(17/3 - 1) - 85/2 x^(-1/2)
f '(x) = 85 x^(14/3) - 85/2 x^(-1/2)
If we factor out 85 x^(-1/2), we have
f '(x) = 85 x^(-1/2) * [x^(31/6) - 1/2]
This equation has two factors:
The first factor (which is in bold above) is really 85 / square root of x. This factor is never 0, but it is undefined when x = 0. Therefore, 0 is a critical number.
The second factor (which is in italics above) is x^(31/6) - 1/2. This is never undefined. If we set it equal to 0, then
x^(31/6) - 1/2 = 0
x^(31/6) = 1/2
Then raise each side to the 6/31 power.
x = (1/2)^(31/6) which is another critical number.
I hope this makes sense! I wish we had a little better way to type math equations here, but we do the best we can!
Jim
