Math help please?

Question

Okay this problem is being impossible, i have no idea what to do. any help is appreciated.

use a differential to approximate sqrt 24.8 . they gave a hint too: Let f(x)= sqrt x, and compare dy with x=25 and dx= -0.2

thanks

Answer 1

If y=sqrt(x), y`=.5/sqrt(x)*dx
Use the hint in the following way: assume x=25 and approximate the square root of 24.8 by calculating the differential when dx=-.2.
dy=.5/5*(-.2)=-.02
Therefore, sqrt(24.8) is approximately 5-.02=4.98.

Answer 2

This is called Newton's approximation, and it is actually very simple.

Remember what a derivative is - it is essentially the slope of the curve at the given point. Imagine a curve near the X axis (I wish I could draw this, it makes it much easier). Draw the tangent of the curve at some point near the X axis. It intercepts the X axis very near where the actual curve itself does, so it's x value is very close to the zero.

It is not hard to get the equation of this line - the slope is simply the derivative of the function at the chosen point (x0, y0), and you know it goes through that point, so knowing a point and the slope it is easy to get the equation of the line.

And, with the equation of a line, it is very easy to solve exactly the point it hits the X axis, just let y = 0 and solve the linear equation. This will give you a new x that is, with any luck, a better approximation than the one you first used. You can then repeat this same technique starting with your new point to get an even better approximation.

In brief, the idea is to get a linear approximation for the function in the neighborhood of 0, then solve that linear equation to get the next approximation.

Answer 3

Yes. Have you differentiated f(x) yet?

You should have got
f'(x) = (1/2)x^(-1/2)

so sub x = 5 to find f'(25).

Then use
f(24.8) = f(25) + dy

where dy = f'(25)*dx with
dx = -0.2 as the hint suggests.

Answer 4

f(x)=25 dx=-0.2

f'(x)=1/2*sqrt (x)
=1/(2*5) * (-0.2)= -0.02

=sqrt of 25 + (-0.02)= 4.98