Chapter 4.5, Problem 69E

To determine

To Show: If the given conditions are imposed on $g$ , then $g\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)$ .

Answer to Problem 69E

It has been shown that if the given conditions are imposed on $g$ , then $g\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)$ .

Explanation of Solution

Given:

A function $y=f\left(x\right)$ which is differentiable at $x=a$ .

The function, $g\left(x\right)=m\left(x-a\right)+c$ , where $m$ and $c$ are constants.

The function, $E\left(x\right)=f\left(x\right)-g\left(x\right)$ .

The conditions to be imposed:

  $E\left(a\right)=0$ ,

  $\underset{x\to a}{\lim }\frac{E\left(x\right)}{x-a}=0$

Concept used:

According to the definition of derivatives:

  $\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}=f^{\prime }\left(a\right)$

Calculation:

It is given that $g\left(x\right)=m\left(x-a\right)+c$ , where $m$ and $c$ are constants.

Put $g\left(x\right)=m\left(x-a\right)+c$ in $E\left(x\right)=f\left(x\right)-g\left(x\right)$ to get,

  $E\left(x\right)=f\left(x\right)-m\left(x-a\right)-c$

Put $x=a$ in the above expression to get,

  $E\left(a\right)=f\left(a\right)-m\left(a-a\right)-c$

Simplifying,

  $E\left(a\right)=f\left(a\right)-c$

Put $E\left(x\right)=f\left(x\right)-m\left(x-a\right)-c$ in $\underset{x\to a}{\lim }\frac{E\left(x\right)}{x-a}$ to get,

  $\underset{x\to a}{\lim }\frac{E\left(x\right)}{x-a}=\underset{x\to a}{\lim }\frac{f\left(x\right)-m\left(x-a\right)-c}{x-a}$

Simplifying,

  $\underset{x\to a}{\lim }\frac{E\left(x\right)}{x-a}=\underset{x\to a}{\lim }\frac{f\left(x\right)-c-m\left(x-a\right)}{x-a}$

On further simplification,

  $\underset{x\to a}{\lim }\frac{E\left(x\right)}{x-a}=\underset{x\to a}{\lim }\left(\frac{f\left(x\right)-c}{x-a}-m\right)$

Finally,

  $\underset{x\to a}{\lim }\frac{E\left(x\right)}{x-a}=\underset{x\to a}{\lim }\frac{f\left(x\right)-c}{x-a}-m$

Now, let the given conditions be imposed.

Then,

  $E\left(a\right)=0$ ,

  $\underset{x\to a}{\lim }\frac{E\left(x\right)}{x-a}=0$

Put $E\left(a\right)=0$ in $E\left(a\right)=f\left(a\right)-c$ to get,

  $f\left(a\right)-c=0$

Simplifying,

  $c=f\left(a\right)$

Put $\underset{x\to a}{\lim }\frac{E\left(x\right)}{x-a}=0$ in $\underset{x\to a}{\lim }\frac{E\left(x\right)}{x-a}=\underset{x\to a}{\lim }\frac{f\left(x\right)-c}{x-a}-m$ to get,

  $\underset{x\to a}{\lim }\frac{f\left(x\right)-c}{x-a}-m=0$

Put $c=f\left(a\right)$ in the above equation to get,

  $\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}-m=0$

As discussed, $\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}=f^{\prime }\left(a\right)$ .

Put this value in the above equation to get,

  $f^{\prime }\left(a\right)-m=0$

Simplifying,

  $m=f^{\prime }\left(a\right)$

Put $m=f^{\prime }\left(a\right)$ and $c=f\left(a\right)$ in $g\left(x\right)=m\left(x-a\right)+c$ to get,

  $g\left(x\right)=f^{\prime }\left(a\right)\left(x-a\right)+f\left(a\right)$

Equivalently,

  $g\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)$

This shows that if the given conditions are imposed on $g$ , then $g\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)$ .

Conclusion:

It has been shown that if the given conditions are imposed on $g$ , then $g\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)$ .