Determine the points at which f'(x)=0 for each of the following functions:
f(x)= x^3+6x^2+1
f'(x)=3x^2+12x
What to do next?
To determine the points at which f'(x) = 0 for the function f(x) = x^3 + 6x^2 + 1, you need to set the derivative f'(x) equal to zero and solve for x.
The derivative of f(x) = x^3 + 6x^2 + 1 is f'(x) = 3x^2 + 12x.
Setting f'(x) equal to zero, we have:
3x^2 + 12x = 0.
Next, we can factor out the common terms from the equation:
3x(x + 4) = 0.
Now, we can apply the zero product property, which states that if a product of two factors is equal to zero, then at least one of the factors must be zero. Thus, we set each factor equal to zero and solve for x:
First factor: 3x = 0
Divide both sides by 3:
x = 0.
Second factor: x + 4 = 0
Subtract 4 from both sides:
x = -4.
Therefore, the points at which f'(x) = 0 for the function f(x) = x^3 + 6x^2 + 1 are x = 0 and x = -4.