# 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.

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}