Question

consider the following function.

$y = x^{3}+3x^{2}-9x + 1$

find $y(x)$.

$y(x)=$

the graph of $y(x)$ has a horizontal tangent line when $y(x)$ is equal to what value?

$y(x)=$

set $y(x)$ equal to the value above and solve for $x$. (enter your answers as a comma - separated list.)

$x=$

find the points on the given curve where the tangent is horizontal.

smaller $x$-value $(x,y)=$

larger $x$-value $(x,y)=$

Explanation:

Step1: Differentiate term - by - term

Using the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have:
$y=x^{3}+3x^{2}-9x + 1$.
$y^{\prime}(x)=\frac{d}{dx}(x^{3})+\frac{d}{dx}(3x^{2})-\frac{d}{dx}(9x)+\frac{d}{dx}(1)$.
$y^{\prime}(x)=3x^{2}+6x - 9$.

Step2: Find the value of $y^{\prime}(x)$ for horizontal tangent

A horizontal tangent line occurs when the slope of the tangent line is 0. So $y^{\prime}(x)=0$.

Step3: Solve for $x$ when $y^{\prime}(x)=0$

Set $3x^{2}+6x - 9 = 0$.
Divide through by 3: $x^{2}+2x - 3=0$.
Factor the quadratic equation: $(x + 3)(x - 1)=0$.
Using the zero - product property, $x+3 = 0$ or $x - 1=0$.
So $x=-3,1$.

Step4: Find the $y$ - values for the horizontal tangent points

When $x=-3$:
$y=(-3)^{3}+3(-3)^{2}-9(-3)+1=-27 + 27+27 + 1=28$.
When $x = 1$:
$y=(1)^{3}+3(1)^{2}-9(1)+1=1 + 3-9 + 1=-4$.

Answer:

$y^{\prime}(x)=3x^{2}+6x - 9$
$y^{\prime}(x)=0$
$x=-3,1$
Smaller $x$ - value: $(x,y)=(-3,28)$
Larger $x$ - value: $(x,y)=(1,-4)$