find critical values for the function sqrt x^2+x-20
Take the second derivative,set to zero, solve for x, then f(x)
I don't know how to do that with the sqrt
sqrt means something to the 1/2 power.
sqrt k= k^(1/2)
derviative of sqrt k= 1/2 * k^-1/2
oh I remember that part but do you do the d/dx of the inside too?
To find the critical values of a function, we need to find the values of x for which the derivative of the function is equal to zero or undefined.
Let's find the derivative of the given function first.
f(x) = sqrt(x^2 + x - 20)
We can rewrite the function as:
f(x) = (x^2 + x - 20)^(1/2)
Now, let's find the derivative f'(x):
f'(x) = (1/2)(x^2 + x - 20)^(-1/2)(2x + 1)
To find the critical values, we need to solve the equation f'(x) = 0 for x:
(1/2)(x^2 + x - 20)^(-1/2)(2x + 1) = 0
Simplifying the equation:
(2x + 1) = 0
Now, solve for x:
2x = -1
x = -1/2
The critical value of the function is x = -1/2.