MHB Help with Functions - Linearization

[vickon]

Let f(x) = \sqrt{x}
Assume that g is function such that
(i) g(c)= c+m(x-1)
(ii) f(1) = g(1), and
(iii) \lim_{{x}\to{1}}\frac{f(x)-g(x)}{x-1}

Answer the following questions. Show all of your work, and explain your reasoning.
(a) What are the constants c and m?
(b) How does g compare with the linearization of f at 1?

For a, I have that the constant c=1, but I'm having trouble determining the constant m. I also am not sure what is required to answer part b.

 

[skeeter]

Let f(x) = \sqrt{x}
Assume that g is function such that

(i) g(c)= c+m(x-1) sure this isn't g(x) = c + m(x-1) ?

(ii) f(1) = g(1), and

(iii) \lim_{{x}\to{1}}\frac{f(x)-g(x)}{x-1} is this limit equal to anything ?

Answer the following questions. Show all of your work, and explain your reasoning.
(a) What are the constants c and m?
(b) How does g compare with the linearization of f at 1?

For a, I have that the constant c=1, but I'm having trouble determining the constant m. I also am not sure what is required to answer part b.

clarification needed above ...

 

[vickon]

Yes, sorry! g(x)=c+m(x-1) and the lim=0

 

[skeeter]

$\displaystyle \lim_{x \to 1} \dfrac{\sqrt{x} - [1+m(x-1)]}{x-1} = 0$

$\displaystyle \lim_{x \to 1} \dfrac{\sqrt{x}-1}{x-1} - \dfrac{m(x-1)}{x-1} = 0$

$\displaystyle \lim_{x \to 1} \dfrac{1}{\sqrt{x}+1} - m = 0 \implies m = \dfrac{1}{2}$linearization of f(x) at x = 1 ...

$L(1) = f(1) + f’(1) \cdot (x-1)$