7. determine all the points on the graph of y = x^3 + x^2 - x + 1 where the tangent is horizontal.
7. determine all the points on the graph of y = x^3 + x^2 - x + 1 where the tangent is horizontal.
Answer
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 $(-1,2)$.