Chapter 2.4, Problem 13E

To determine

To find:The slope of secant line connecting two points of the curve of given quadratic function, and the slope tangent line as a function of x in both differential and prime notation.

Answer to Problem 13E

The slope of secant to the curve $f\left(x\right)=4-x^2$ connecting points $\left(x,f\left(x\right)\right)$ and $\left(x+△x,f\left(x+△x\right)\right)$ is $-\left(2+△x\right)$ , and the slope of tangent at point $\left(x,y\right)$ is

  $\frac{df}{dx}=-2x$ (Differential form), or $f^{\prime }\left(x\right)=-2x$ (Prime notation).

Explanation of Solution

Given information:

The given quadratic function is $f\left(x\right)=4-x^2$ .

Method/Formulaused:

The slope of a line is tangent to the angle, line makes with x -axis. If $y=f\left(x\right)$ is a given curve then differential form of the slope of a line is $\frac{df}{dx}$ and the prime notation form is $f^{\prime }\left(x\right)$

Calculations:

Let us consider two neighboring points $P\left(x,f(x)\right)$ and $Q\left(x+△x,f(x+△x)\right)$ on the curve of a quadratic function $f\left(x\right)$ as shown in figure here.

  

The line PQ is a secant line connecting points $P\left(x,f(x)\right)$ and $Q\left(x+△x,f(x+△x)\right)$ , and making an angle $\theta$ (say) with x -axis.

In figure, the coordinates of point N are $\left(x+△x,f(x)\right)$ . Therefore, $PN=\left(x+△x\right)-\left(x\right)$ and $NQ=f\left(x+△x\right)-f\left(x\right)$ .

Hence, the slope of secant PQ is

  $m\text{=tanθ}=\frac{NQ}{PN}=\frac{f\left(x+△x\right)-f\left(x\right)}{\left(x+△x\right)-\left(x\right)}$  ... (1)

For the given function, $f\left(x\right)=4-x^2$$m=\frac{\left\{4-{\left(x+△x\right)}^2\right\}-\left\{4-x^2\right\}}{△x}=\frac{x^2-{\left(x+△x\right)}^2}{△x}=\frac{-2x\cdot △x-{\left(△x\right)}^2}{△x}$

  $\Rightarrow m=-\left(2x+△x\right)$

We may write $\frac{f\left(x+△x\right)-f\left(x\right)}{△x}=\frac{△f}{△x}$ .

Thus, the slope of secant to the curve $f\left(x\right)=4-x^2$ connecting points $\left(x,f\left(x\right)\right)$ and $\left(x+△x,f\left(x+△x\right)\right)$ is $-\left(2+△x\right)$ .

The secant PQbecomes tangent when point Q coincides with point P that is when $△x\to 0$ .

The slope of tangent is the limit of the slope of secant as $△x\to 0$ .

Now, taking limit as $△x\to 0$ , $m=\underset{△x\to 0}{\lim }\frac{△f}{△x}=\frac{df}{dx}$

  $\Rightarrow \frac{df}{dx}=\underset{△x\to 0}{\lim }\frac{△f}{△x}=\underset{△x\to 0}{\lim }\left\{-\left(2x+△x\right)\right\}=-2x$ .

Thus, the slope of the tangent at point $\left(x,y\right)$ is

  $\frac{df}{dx}=-2x$

(Differential form), or

  $f^{\prime }\left(x\right)=-2x$

(Prime notation)

Conclusion:

The slope of secant to the curve $f\left(x\right)=4-x^2$ connecting points $\left(x,f\left(x\right)\right)$ and $\left(x+△x,f\left(x+△x\right)\right)$ is $-\left(2+△x\right)$ , and the slope of tangent at point $\left(x,y\right)$ is

  $\frac{df}{dx}=-2x$ (Differential form), or $f^{\prime }\left(x\right)=-2x$ (Prime notation).