Final answer:
To find the values of x for which f'(x)=0, we need to find the derivative of f(x) and set it equal to 0. Given f(x)=(2x-2)(x²-3), we first find f'(x) using the chain rule. We then set f'(x) equal to 0 and solve for x.
Step-by-step explanation:
To find the values of x for which f'(x)=0, we need to find the derivative of f(x) and set it equal to 0. Given f(x)=(2x-2)x²-3, we need to first find f'(x) by using the chain rule. Taking the derivative, we get f'(x) = (2x-2)x²-3 * (2 + 2x(2x-2)ln(2x-2)). Now, we set f'(x) equal to 0 and solve for x.
Starting with the equation f'(x) = (2x-2)x²-3 * (2 + 2x(2x-2)ln(2x-2)) = 0, we have two possible cases: (2x-2)x²-3 = 0 and (2 + 2x(2x-2)ln(2x-2)) = 0.
For the first case, (2x-2)x²-3 = 0, we can set 2x-2 = 0 and solve for x. This gives us x = 1.
For the second case, (2 + 2x(2x-2)ln(2x-2)) = 0, we solve the equation using numerical methods or graphing to find the approximate value of x.