Chapter 2.4, Problem 14E

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 14E

The slope of secant to the curve $g\left(x\right)=x+2x^2$ connecting points $\left(x,g\left(x\right)\right)$ and $\left(x+△x,g\left(x+△x\right)\right)$ is $1+4x+△x$ , and the slope of tangent at point $\left(x,y\right)$ is

  $\frac{dg}{dx}=1+4x$ (Differential form), or $f^{\prime }\left(x\right)=1+4x$ (Prime notation).

Explanation of Solution

Given information:

The given quadratic function is $g\left(x\right)=x+2x^2$ .

Method/Formulaused:

The slope of a line is tangent of the angle, the line makes with x -axis. Let $y=f\left(x\right)$ is a given curve, then the differential form of the slope of tangent 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 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 the 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, $g\left(x\right)=x+2x^2$

  $m\text{=tanθ}=\frac{NQ}{PN}=\frac{g\left(x+△x\right)-g\left(x\right)}{\left(x+△x\right)-\left(x\right)}$$m=\frac{\left\{\left(x+△x\right)+2{\left(x+△x\right)}^2\right\}-\left\{x+2x^2\right\}}{△x}$

  $\Rightarrow m=\frac{\left(x+△x\right)+2{\left(x+△x\right)}^2-\left(x+2x^2\right)}{△x}=\frac{x+△x+2x^2+4x△x+2{\left(△x\right)}^2-x-2x^2}{△x}$

  $\Rightarrow m=\frac{△x+4x\cdot △x+{\left(△x\right)}^2}{△x}=1+4x+△x$

Thus, the slope of secant to the curve $g\left(x\right)=x+2x^2$ connecting points $\left(x,g\left(x\right)\right)$ and $\left(x+△x,g\left(x+△x\right)\right)$ is $m=1+4x+△x$ .

Further, we may write $m=\frac{g\left(x+△x\right)-g\left(x\right)}{△x}=\frac{△g}{△x}$ .

The secant PQbecomes a tangent at pointP 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 $△x\to 0$ , $m=\underset{△x\to 0}{\lim }\frac{△g}{△x}=\frac{dg}{dx}$

  $\Rightarrow \frac{dg}{dx}=\underset{△x\to 0}{\lim }\frac{△g}{△x}=\underset{△x\to 0}{\lim }\left\{1+4x+△x\right\}=1+4x+0=1+4x$ .

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

  $\frac{dg}{dx}=1+4x$

(Differential form), or

  $g^{\prime }\left(x\right)=1+4x$

(Prime notation)

Conclusion:

The slope of secant to the curve $g\left(x\right)=x+2x^2$ connecting points $\left(x,g\left(x\right)\right)$ and $\left(x+△x,g\left(x+△x\right)\right)$ is $1+4x+△x$ , and the slope of tangent at point $\left(x,y\right)$ is

  $\frac{dg}{dx}=1+4x$ (Differential form), or $f^{\prime }\left(x\right)=1+4x$ (Prime notation).