Copyright

The linearisation of the function f at the point x = 2 is L(x) = 5x - 8. Let K be the...

Question:

The linearisation of the function {eq}f {/eq} at the point {eq}x = 2 {/eq} is {eq}L(x) = 5x - 8 {/eq}. Let {eq}K {/eq} be the linearisation of the function {eq}u(x) = \frac{f(x)}{x} {/eq} at {eq}x = 2 {/eq}. Find {eq}K {/eq}.

Linearisation

Using the method of linearisation we perform the linear approximation of the inputs and the corresponding output. We have all the necessary information to perform the linearization and calculate the value of K.

Answer and Explanation:


The function we have is,

{eq}\displaystyle L(x)=5x-8 {/eq}

We have to find the linearization at {eq}x=2 {/eq}

So,

{eq}\displaystyle f(x)=f(2)+f'(x-2) {/eq}

{eq}\displaystyle \Rightarrow f(x)=f'(2)x+\left ( f(2)-f'(2) \right ) {/eq}

Equating

{eq}\displaystyle L(x)=5x-8 {/eq}

Now,

{eq}f'(2)=5 {/eq}

{eq}f(2)-2f'(2)=-8 {/eq}

{eq}f(2)-2\times 5=-8 {/eq}

{eq}f(2)=2 {/eq}


Now,

{eq}\displaystyle u(x)=\frac{f(x)}{x} {/eq}

{eq}\displaystyle u(2)=\frac{f(2)}{2}=\frac{2}{2}=1 {/eq}

Also,

{eq}\displaystyle u'(x)=\frac{f'(x)}{x}-\frac{f(x)}{x^2} {/eq}

{eq}\displaystyle u'(2)=\frac{f'(2)}{2}-\frac{2}{4} {/eq}

{eq}\displaystyle u'(2)=\frac{f'(2)}{2}-\frac{2}{4} {/eq}

{eq}\displaystyle u'(2)=\frac{5}{2}-\frac{2}{4} {/eq}

{eq}\displaystyle u'(2)=2 {/eq}


So, the linearization of {eq}u(x) {/eq} is,

{eq}\displaystyle u(x)=u(2)+u'(2)(x-2) {/eq}

{eq}\displaystyle u(x)=1+2(x-2) {/eq}

Thus, the value of K is:

{eq}\displaystyle \boxed{\displaystyle K(x)=2x-3} {/eq}


Learn more about this topic:

Loading...
Linearization of Functions

from Math 104: Calculus

Chapter 10 / Lesson 1
7.4K

Related to this Question

Explore our homework questions and answers library