Chapter 4, Problem 10RWDT

To determine

To find: Conclude about $a,b,c$

Answer to Problem 10RWDT

The conclusion that derived from the linearization is that $b=m,c=0$ .

Explanation of Solution

Given information:

  $y=mx$ is the linearization at the origin of parabola $y=a{x}^2+bx+c$ .

Calculation:

The linearization of a function gives the equation of tangent line. The equation of the tangent line of the parabola is given as $y=mx$ .

In order to find the relation of $a,b$ and $c$ , it is required to find the linearization of parabola at the origin.

Here the formula of linearization is $L\left(x\right)=f\left(a\right)+f\text{'}\left(a\right)\left(x-a\right)$

Here $a$ is given origin that is $x=0$ and $f(x)=a{x}^2+bx+c$

Find the value of $y=a{x}^2+bx+c$ at $x=0$ .

  $\begin{array}{l}f(0)=a\cdot 0+b\cdot 0+c\\ f\left(0\right)=c\end{array}$

Now find $f\text{'}(x)$ by differentiating $f\left(x\right)$ with respect to $x$ :

  $\begin{array}{l}f\text{'}\left(x\right)=\frac{d}{dx}\left(a{x}^2+bx+c\right)\\ f\text{'}\left(x\right)=\frac{d}{dx}\left(a{x}^2\right)+\frac{d}{dx}\left(bx\right)+\frac{d}{dx}\left(d\right)\\ f\text{'}\left(x\right)=2ax+b\left[\because \frac{d}{dx}\left(x^n\right)=n{x}^{n-1}\right]\end{array}$

Now substitute $x=0$ in $f\text{'}\left(x\right)=2ax+b$ gives:

  $\begin{array}{l}f\text{'}\left(0\right)=2a\cdot 0+b\\ f\text{'}\left(0\right)=b\end{array}$

Substitute the value of $f\left(0\right)$ and $f\text{'}\left(0\right)$ in $L\left(x\right)=f\left(a\right)+f\text{'}\left(a\right)\left(x-a\right)$ :

Here $a=0$ .

  $L\left(0\right)=f\left(0\right)+f\text{'}\left(0\right)\left(x-0\right)$

  $\begin{array}{l}L\left(0\right)=c+b\left(x-0\right)\\ L\left(0\right)=c+bx\\ y=c+bx\end{array}$

This is the equation of tangent line.

Now compare $y=c+bx$ with given line $y=mx$ gives that $b=m,c=0$ .