Question

1. determine all the points on the graph of y = x^3 + x^2 - x + 1 where the tangent is horizontal.

Explanation:

Step1: Find the derivative

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

Step2: Set the derivative equal to 0

Since the slope of a horizontal tangent is 0, we set $y' = 0$. So, $3x^{2}+2x - 1=0$.

Step3: Solve the quadratic equation

Factor the quadratic equation: $3x^{2}+2x - 1=(3x - 1)(x + 1)=0$. Then, $3x-1 = 0$ gives $x=\frac{1}{3}$, and $x + 1=0$ gives $x=-1$.

Step4: Find the corresponding y - values

When $x=\frac{1}{3}$, $y=(\frac{1}{3})^{3}+(\frac{1}{3})^{2}-\frac{1}{3}+1=\frac{1}{27}+\frac{1}{9}-\frac{1}{3}+1=\frac{1 + 3-9 + 27}{27}=\frac{22}{27}$.
When $x=-1$, $y=(-1)^{3}+(-1)^{2}-(-1)+1=-1 + 1+1 + 1=2$.

Answer:

The points are $(\frac{1}{3},\frac{22}{27})$ and $(-1,2)$.