Chapter 2.4, Problem 20E

(a)

To determine

To find: The slope of the curve $y=\frac{2}{x}$ at $x=a$.

Answer to Problem 20E

The slope of the curve $y=\frac{2}{x}$ at $x=a$ is equal to $-\frac{2}{a^2}$.

Explanation of Solution

Given information:

The curve is $y=\frac{2}{x}$ and point is $x=a$.

Calculation:

The formula for the slope of the curve $f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$.

The function is given as $f\left(x\right)=\frac{2}{x}$.

Substitute $a+h$ for $x$ in the function.

  $f\left(a+h\right)=\frac{2}{a+h}$

Substitute $a$ for $x$ in the function.

  $f\left(a\right)=\frac{2}{a}$

Substitute $\frac{2}{a+h}$ for $f\left(a+h\right)$ and $\frac{2}{a}$ for $f\left(a\right)$ in the formula for slope.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0}{\lim }\frac{\frac{2}{a+h}-\frac{2}{a}}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{2a-2\left(a+h\right)}{a\left(a+h\right)}}{h}\\ =\underset{h\to 0}{\lim }\frac{-2h}{ah\left(a+h\right)}\\ =\underset{h\to 0}{\lim }\left(\frac{-2}{a\left(a+h\right)}\right)\end{array}$

Further simplify.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\frac{-2}{a\left(a+0\right)}\\ =-\frac{2}{a^2}\end{array}$

Therefore, the slope of the curve $y=\frac{2}{x}$ at $x=a$ is $-\frac{2}{a^2}$.

(b)

To determine

To find: The change in the tangent at $x=a$ as the value of $a$ changes.

Answer to Problem 20E

The slope of the tangent decreases as the value of $a$ increases.

Explanation of Solution

Given information:

The curve is $y=\frac{2}{x}$ and point is $x=a$.

Calculation:

As calculated in part (a), the slope of the curve $y=\frac{2}{x}$ at $x=a$ is equal to $-\frac{2}{a^2}$.

So, the slope of the curve is inversely proportional to the square of $a$ and the tangent is steeper near origin.

Therefore, the slope of the tangent decreases as the value of $a$ increases.