1. ## Weird problem

Show that for small values of h, sqroot(4+h) is approx equal to 2 + (h/4)

2. Originally Posted by mightymax
Show that for small values of h, sqroot(4+h) is approx equal to 2 + (h/4)
This is a linearzation problem.

Consider the function,
$\sqrt{4+h}$
And its tangent line at point $h=0$.
We know that a small change in the actual function $\Delta h$ is approximately equal to a small change in the tangent line $dh$.
That is,
$\Delta h\approx dh$
But the derivative is,
(Hint: $\sqrt{4+h}=(4+h)^{1/2}$)

$\frac{4+h}{2}=2+(h/2)$
(The reason why we look at the derivative is because the derivative is the tangent line. So a small change in the derivative is a small change in the function).

3. Thanks for the help

but I dont really understand how to answer the question. And do I look at the derivative of the sqroot(4+h) function? and if so what do i do next?

4. Originally Posted by mightymax
Show that for small values of h, sqroot(4+h) is approx equal to 2 + (h/4)
Hello,

I'll try to explain what TPH meant.

You've got a function $v(h)=\sqrt{4+h}$ which gives for any h the value of the square-root. You know that v(0) = 2.

Now consider the tangent line at this point. For small amounts of h the tangent line and the curve of the function v are practivally the same. (see attachment. I've enlarged a small surrounding of (0, v(0)))

To calculate the tangent line you need the slope of the line which is the same as the gradient of v for h = 0:
$v'(h)=\frac{1}{2}\cdot \frac{1}{\sqrt{4+h}}$. Thus $v'(0)=\frac{1}{4}$

Therefore the tangent line has the equation:
$t(h)=\frac{1}{4}\cdot h +2$

Because t(h) is the same as v(h) for small values of h, you can write (or calculate): $\sqrt{4+h}\approx \frac{1}{4}\cdot h +2$

EB