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,
And its tangent line at point .
We know that a small change in the actual function is approximately equal to a small change in the tangent line .
That is,
But the derivative is,
(Hint: )
(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).
Hello,
I'll try to explain what TPH meant.
You've got a function 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:
. Thus
Therefore the tangent line has the equation:
Because t(h) is the same as v(h) for small values of h, you can write (or calculate):
EB