Find all relative extrema of the function. `f(x)=x-6 sqrtx` , x>0
this is what I've come up with 1-(6x^1/2)=0
I have more but am not sure it is right.
Thankyou for your help.
To determine the relative extrema, determine f'(x).
`f'(x)= d/(dx) x - d/(dx)6sqrtx`
Let's take the derivative of each term.
For the first term,
>>> `d/(dx)x = 1`
For the second term, apply the power rule of derivative which is `d/(dx) (cx^n) = c*nx^(n-1)` .
>>> `d/(dx) 6sqrtx= d/(dx)6x^(1/2)=6*1/2x^(-1/2) = 3/x^(1/2)`
So, f'(x) is:
`f'(x) = 1-3/x^(1/2)`
Set f'(x) equal to zero. And solve for x.
To isolate x, square both sides.
Then, substitute value of x to f(x).
Hence, at `xgt0`, the relative extrema of `f(x)=x-6sqrtx` is `(9,-9)`.