선형 근사 계산기

Your input: find the linear approximation to $$$f(x)=\sqrt{x}$$$ at $$$x_0=4$$$.

A linear approximation is given by $$$L(x)\approx f(x_0)+f^{\prime}(x_0)(x-x_0)$$$.

We are given that $$$x_0=4$$$.

Firstly, find the value of the function at the given point: $$$y_0=f(x_0)=2$$$.

Secondly, find the derivative of the function, evaluated at the point: $$$f^{\prime}\left(4\right)$$$.

Find the derivative: $$$f^{\prime}\left(x\right)=\frac{1}{2 \sqrt{x}}$$$ (steps can be seen here).

Next, evaluate the derivative at the given point to find slope.

$$$f^{\prime}\left(4\right)=\frac{1}{4}$$$.

Plugging the values found, we get that $$$L(x)\approx 2+\frac{1}{4}\left(x-\left(4\right)\right)$$$.

Or, more simply: $$$L(x)\approx \frac{1}{4} x+1$$$.

Answer: $$$L(x)\approx \frac{1}{4} x+1 =0.25x+1$$$.