Question

this is section 2.5 problem 56: find the points on the curve f(x)=2x^3 - 3x^2+5 where the tangent line is horizontal. list your answer in ascending order of x. answer: ( , ) and ( , )

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=2x^{3}-3x^{2}+5$ using the power - rule $(x^n)' = nx^{n - 1}$ is $f'(x)=6x^{2}-6x$.

Step2: Set the derivative equal to zero

Since the slope of a horizontal tangent line is 0, we set $f'(x) = 0$. So, $6x^{2}-6x=0$.
Factor out $6x$: $6x(x - 1)=0$.

Step3: Solve for x

Set each factor equal to zero:
If $6x=0$, then $x = 0$.
If $x - 1=0$, then $x = 1$.

Step4: Find the corresponding y - values

When $x = 0$, $f(0)=2(0)^{3}-3(0)^{2}+5=5$.
When $x = 1$, $f(1)=2(1)^{3}-3(1)^{2}+5=2 - 3 + 5=4$.

Answer:

$(0,5)$ and $(1,4)$