Let f(x) = (2)sqrt(x). If f(c)=f'(c) find the value of c
f(x) = 2√x = 2x^(1/2) ----> f(c) = 2√c
f'(x) = (1/2)(2)x^(-1/2) = 1/√x , ------> f'(c) = 1/√c
they are supposed to be equal, so
2√c = 1/√c
multiply both sides by √c
2c = 1
etc
To find the value of c such that f(c) = f'(c), we need to set up the equation and solve for c.
First, let's find the derivative of f(x):
f(x) = (2)sqrt(x)
Using the power rule and the chain rule, we can differentiate f(x) as follows:
f'(x) = 2 * (1/2) * x^(-1/2)
= x^(-1/2)
= 1/sqrt(x)
Now, let's set up the equation f(c) = f'(c) and solve for c:
(2)sqrt(c) = 1/sqrt(c)
Next, let's square both sides of the equation to eliminate the square root:
4c = 1/c
Now, let's multiply both sides of the equation by c to get rid of the fraction:
4c^2 = 1
Finally, divide both sides of the equation by 4 to solve for c:
c^2 = 1/4
Taking the square root of both sides, we have:
c = ±1/2
Therefore, the two possible values of c such that f(c) = f'(c) are c = 1/2 and c = -1/2.