Question

find the points on the curve y = x^3 + 3x^2 - 9x + 8 where the tangent is horizontal. smaller x - value (x, y) = ( ) larger x - value (x, y) = ( )

Explanation:

Step1: Find the derivative

The derivative of $y = x^{3}+3x^{2}-9x + 8$ using the power - rule $(x^n)'=nx^{n - 1}$ is $y'=3x^{2}+6x - 9$.

Step2: Set the derivative equal to 0

Since the slope of a horizontal tangent is 0, we set $y' = 0$. So, $3x^{2}+6x - 9=0$. Divide through by 3 to get $x^{2}+2x - 3=0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $x^{2}+2x - 3=(x + 3)(x - 1)=0$. Then, $x=-3$ or $x = 1$.

Step4: Find the y - values

When $x=-3$, $y=(-3)^{3}+3(-3)^{2}-9(-3)+8=-27 + 27+27 + 8=35$.
When $x = 1$, $y=1^{3}+3\times1^{2}-9\times1 + 8=1+3-9 + 8=3$.

Answer:

smaller $x$-value $(x,y)=(-3,35)$
larger $x$-value $(x,y)=(1,3)$