Question

a. for f(x) = 8/x, find an equation for the secant line through the points where x = 1 and x = 8. the equation for the secant line is y=-1/5x + 13/5. b. for f(x) = 8/x, find an equation for the line tangent to the curve when x = 1.

Explanation:

Step1: Recall secant - line formula

The formula for the slope of the secant line of a function $y = f(x)$ through the points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is $m=\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Given $f(x)=\frac{8}{x}$, when $x_1 = 1$, $f(1)=\frac{8}{1}=8$; when $x_2 = 8$, $f(8)=\frac{8}{8}=1$. Then $m=\frac{1 - 8}{8 - 1}=\frac{-7}{7}=-1$. Using the point - slope form $y - y_1=m(x - x_1)$ with the point $(1,8)$ and $m=-1$, we have $y-8=-1(x - 1)$, which simplifies to $y=-x + 9$.

Step2: Recall tangent - line formula

The derivative of $y = f(x)=\frac{8}{x}=8x^{-1}$ using the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$ is $f^\prime(x)=-8x^{-2}=-\frac{8}{x^{2}}$. When $x = 1$, the slope of the tangent line $m=f^\prime(1)=-8$. Using the point - slope form $y - y_1=m(x - x_1)$ with the point $(1,8)$ and $m=-8$, we have $y - 8=-8(x - 1)$, which simplifies to $y=-8x+16$.

Answer:

a. $y=-x + 9$
b. $y=-8x + 16$