# Help with Functions - Linearization

• MHB
• vickon
In summary: L(1) = \sqrt{1} + \dfrac{1}{2}(x-1) = \dfrac{1}{2}x + \dfrac{1}{2}$In summary, the function f(x) =$\sqrt{x}$and the function g(x) = c + m(x-1) are given. It is known that g(1) = f(1), and the limit as x approaches 1 of the difference quotient of f(x) and g(x) is equal to 0. Solving vickon Let f(x) = \sqrt{x} Assume that g is function such that (i) g(c)= c+m(x-1) (ii) f(1) = g(1), and (iii) \lim_{{x}\to{1}}\frac{f(x)-g(x)}{x-1} Answer the following questions. Show all of your work, and explain your reasoning. (a) What are the constants c and m? (b) How does g compare with the linearization of f at 1? For a, I have that the constant c=1, but I'm having trouble determining the constant m. I also am not sure what is required to answer part b. vickon said: Let f(x) = \sqrt{x} Assume that g is function such that (i) g(c)= c+m(x-1) sure this isn't g(x) = c + m(x-1) ? (ii) f(1) = g(1), and (iii) \lim_{{x}\to{1}}\frac{f(x)-g(x)}{x-1} is this limit equal to anything ? Answer the following questions. Show all of your work, and explain your reasoning. (a) What are the constants c and m? (b) How does g compare with the linearization of f at 1? For a, I have that the constant c=1, but I'm having trouble determining the constant m. I also am not sure what is required to answer part b. clarification needed above ... Yes, sorry! g(x)=c+m(x-1) and the lim=0$\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)\$

## What is linearization?

Linearization is the process of approximating a nonlinear function with a linear function. It involves finding the tangent line at a specific point on the curve and using it to approximate the function's behavior near that point.

## Why is linearization useful?

Linearization is useful because it allows us to simplify complex nonlinear functions and make them easier to analyze. It also helps us approximate the behavior of a function near a specific point without having to use complicated calculations.

## How do you find the linearization of a function?

The linearization of a function can be found by taking the first derivative of the function, plugging in the specific point of interest, and using the point-slope form to create the equation of the tangent line.

## What is the difference between linearization and linear regression?

Linearization and linear regression are two different methods used to approximate nonlinear functions. Linearization involves finding the tangent line at a specific point, while linear regression involves finding the best-fit line for a set of data points.

## Can linearization be used for all types of functions?

No, linearization can only be used for functions that are differentiable at the point of interest. If a function is not differentiable at a specific point, linearization cannot be used to approximate its behavior at that point.

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