Respuesta :
Answer:
[tex]L(x)=\frac{1}{8}x+2\\ L(16.4)=4.05[/tex]
Step-by-step explanation:
The equation for a tangent line of f(x) in the point (a,f(a)) can be calculated as:
L(x) = f(a) + f'(a)(x-a)
Where L(x) is also call a linear approximation and f'(a) is the value of the derivative of f(x) in (a,f(a)).
So, the derivative of f(x) is:
[tex]f(x)=\sqrt{x} \\f'(x)=\frac{1}{2\sqrt{x} }[/tex]
Then, to find the linear approximation we are going to use the point (16, f(16)). So a is 16 and f(a) and f'(a) are calculated as:
[tex]f(16)=\sqrt{16}=4\\f'(16)=\frac{1}{2\sqrt{16} }=\frac{1}{8}[/tex]
Then, replacing the values, we get that the equation of the tangent line in (16,4) is:
[tex]L(x)=4+\frac{1}{8}(x-16)\\L(x) = \frac{1}{8}x+2[/tex]
Finally, the approximation for [tex]\sqrt{16.4}[/tex] is:
[tex]L(16.4)=\frac{1}{8}(16.4)+2\\ L(16.4)=4.05[/tex]
