Answer:
[tex]\displaystyle L(x)=1+\frac{1}{4}x[/tex]
Step-by-step explanation:
Linearization
It consists in finding a linear approximate function around certain point x=a of a non linear function. The equation used to linearize a function f is given by
[tex]L(x)=f(a)+f'(a)(x-a)[/tex]
Where f'(a) is the first derivative of f at x=a
The questions gives us this function
[tex]f(x)=\sqrt{x}[/tex]
And the point a=4. Let's compute f(a)
[tex]f(4)=\sqrt{4}=2[/tex]
We are using only the positive root. Now we compute the derivative
[tex]\displaystyle f'(x)=\frac{1}{2\sqrt{x}}[/tex]
[tex]\displaystyle f'(4)=\frac{1}{2\sqrt{4}}[/tex]
[tex]\displaystyle f'(4)=\frac{1}{4}[/tex]
Replacing the values into the equation
[tex]\displaystyle L(x)=2+\frac{1}{4}(x-4)[/tex]
[tex]\displaystyle L(x)=1+\frac{1}{4}x[/tex]
