66. find the points on the graph of f(x) = x² + 1 when the slope is (a) - 1, (b) 0, and (c) 1.

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=x^{2}+1$ using the power - rule $(x^n)^\prime=nx^{n - 1}$ is $f^\prime(x)=2x$. The derivative gives the slope of the tangent line to the graph of the function at a given point.

Step2: Solve for $x$ when the slope is - 1

Set $f^\prime(x)=-1$. So, $2x=-1$. Solving for $x$, we get $x =-\frac{1}{2}$. Substitute $x =-\frac{1}{2}$ into the original function $f(x)$: $f(-\frac{1}{2})=(-\frac{1}{2})^{2}+1=\frac{1}{4}+1=\frac{5}{4}$. The point is $(-\frac{1}{2},\frac{5}{4})$.

Step3: Solve for $x$ when the slope is 0

Set $f^\prime(x)=0$. So, $2x = 0$. Solving for $x$, we get $x = 0$. Substitute $x = 0$ into the original function $f(x)$: $f(0)=0^{2}+1 = 1$. The point is $(0,1)$.

Step4: Solve for $x$ when the slope is 1

Set $f^\prime(x)=1$. So, $2x=1$. Solving for $x$, we get $x=\frac{1}{2}$. Substitute $x=\frac{1}{2}$ into the original function $f(x)$: $f(\frac{1}{2})=(\frac{1}{2})^{2}+1=\frac{1}{4}+1=\frac{5}{4}$. The point is $(\frac{1}{2},\frac{5}{4})$.

Answer:

(a) $(-\frac{1}{2},\frac{5}{4})$ (b) $(0,1)$ (c) $(\frac{1}{2},\frac{5}{4})$