Question

To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equation of the tangent line at the point (4, 2), we know that (4, 2) is a point on the line. So we just need to find its slope. The slope of a tangent line to f(x) at x = a can be found using the formula

Answer 1

Answer:

$x-4y+4=0$

$f(x)=\sqrt x$ and x=4

Step-by-step explanation:

We are given that a curve

$y=\sqrt x$

We have to find the equation of tangent at point (4,2) on the given curve.

Let y=f(x)

Differentiate w.r.t x

$f'(x)=\frac{dy}{dx}=\frac{1}{2\sqrt x}$

By using the formula $\frac{d(\sqrt x)}{dx}=\frac{1}{2\sqrt x}$

Substitute x=4

Slope of tangent

$m=f'(x)=\frac{1}{2\sqrt 4}=\frac{1}{2\times 2}=\frac{1}{4}$

In given question

$m=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$

$\frac{1}{4}=\lim_{x\rightarrow 4}\frac{f(x)-f(4)}{x-4}$

By comparing we get a=4

Point-slope form

$y-y_1=m(x-x_1)$

Using the formula

The equation of tangent at point (4,2)

$y-2=\frac{1}{4}(x-4)$

$4y-8=x-4$

$x-4y-4+8=0$

$x-4y+4=0$