Chapter 12.3, Problem 9E

To determine

Sketch a graph of the given and the tangent line at point $(1,f(1))$ , find the approximate slope of the tangent line.

Answer to Problem 9E

The approximate slope of the curve at point $(x,y)$ is $\text{Slope}=2$

Explanation of Solution

Given:

The function given is:

  $f(x)=x^2-2$

Consider the function:

  $f(x)=x^2-2$

Since, the tangent is drawn to be on the point $(1,f(1))$ .

  $\begin{array}{cc}f(1)& =1^2-2\\ & =1-2\\ & =-1\end{array}$

The point (1,-1) as marked in the graph below. Now, draw the tangent at (1,-1).

  

By the above graph the slope at point $(1,-1)$ will be as,

  $\text{Slope}=\frac{\text{change in y}}{\text{change in x}}$

Now, from the above graph, the tangent crosses the $x$ axis at $\frac{3}{2}$ and $y$ at $-3$ .

So, the change in y is equal to final position- initial position

  $\Delta y=0-(-3)=3$

Similarly the change in x

  $\Delta x=\frac{3}{2}-0=\frac{3}{2}$

Hence, by the above formula of slope of the tangent

  $\begin{array}{l}\text{Slope}=\frac{3}{\frac{3}{2}}\\ =2\end{array}$

Therefore, the approximate slope of the curve at point $(1,-1)$ is $\text{Slope}=2$