Answer :
Answer:
[tex]\displaystyle y=5+\frac{x}{10}[/tex]
[tex]\displaystyle y=10+\frac{(x-75)}{20}[/tex]
Step-by-step explanation:
Linearization
It consists of finding an approximately linear function that behaves as close as possible to the original function near a specific point.
Let y=f(x) a real function and (a,f(a)) the point near which we want to find a linear approximation of f. If f'(x) exists in x=a, then the equation for the linearization of f is
[tex]y=f(x)=f(a)+f'(a)(x-a)[/tex]
Let's find the linearization for the function
[tex]y=\sqrt{25+x}[/tex]
at (0,5) and (75,10)
Computing f'(x)
[tex]\displaystyle f'(x)=\frac{1}{2\sqrt{25+x}}[/tex]
At x=0:
[tex]\displaystyle f'(0)=\frac{1}{2\sqrt{25+0}}=\frac{1}{10}[/tex]
We find f(0)
[tex]f(0)=\sqrt{25+0}=5[/tex]
Thus the linearization is
[tex]\displaystyle y=f(0)+f'(0)(x-0)=5+\frac{1}{10}x[/tex]
[tex]\displaystyle y=5+\frac{x}{10}[/tex]
Now at x=75:
[tex]\displaystyle f'(75)=\frac{1}{2\sqrt{25+75}}=\frac{1}{20}[/tex]
We find f(75)
[tex]f(75)=\sqrt{25+75}=10[/tex]
Thus the linearization is
[tex]\displaystyle y=f(75)+f'(75)(x-75)=10+\frac{1}{20}(x-75)[/tex]
[tex]\displaystyle y=10+\frac{(x-75)}{20}[/tex]