Question

estimating slope in exercises 5 and 6, use the graph to estimate the slope of the tangent line to y = x^n at the point (1, 1). verify your answer analytically. to print an enlarged copy of the graph, go to mathgraphs.com. 5. (a) y = x^(1/2) (b) y = x^3 6. (a) y = x^(-1/2) (b) y = x^(-1)

Explanation:

Step1: Recall the power - rule for differentiation

The power - rule states that if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. The slope of the tangent line to the curve $y = f(x)$ at the point $(x_0,y_0)$ is given by $f^\prime(x_0)$.

Step2: Calculate the derivative for part (a) of exercise 5

For $y = x^{\frac{1}{2}}$, by the power - rule, $y^\prime=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}$. Evaluate at $x = 1$: $y^\prime(1)=\frac{1}{2}(1)^{-\frac{1}{2}}=\frac{1}{2}$.

Step3: Calculate the derivative for part (b) of exercise 5

For $y = x^{3}$, by the power - rule, $y^\prime = 3x^{3 - 1}=3x^{2}$. Evaluate at $x = 1$: $y^\prime(1)=3(1)^{2}=3$.

Step4: Calculate the derivative for part (a) of exercise 6

For $y = x^{-\frac{1}{2}}$, by the power - rule, $y^\prime=-\frac{1}{2}x^{-\frac{1}{2}-1}=-\frac{1}{2}x^{-\frac{3}{2}}$. Evaluate at $x = 1$: $y^\prime(1)=-\frac{1}{2}(1)^{-\frac{3}{2}}=-\frac{1}{2}$.

Step5: Calculate the derivative for part (b) of exercise 6

For $y = x^{-1}$, by the power - rule, $y^\prime=-1x^{-1 - 1}=-x^{-2}$. Evaluate at $x = 1$: $y^\prime(1)=-(1)^{-2}=-1$.

Answer:

1. (a) $\frac{1}{2}$

(b) $3$

1. (a) $-\frac{1}{2}$

(b) $-1$