# Suppose that I don't have a formula for g(x) but I know that g(1) = 3 and g'(x) = sqrt(x^2+15) for all x. How do I use a linear approximation to estimate g(0.9) and g(1.1)?

Feb 3, 2015

Bear with me a bit, but it involves the slope-intercept equation of a line based on the 1st derivative... And I would like to lead you to the way to do the answer, not just give you the answer...

Okay, before I get to the answer, I'll let you in on the (somewhat) humorous discussion my office mate and I just had...

Me: "Okay, waitasec... You don't know g(x), but you know the derivative is true for all (x)... Why do you want to do a linear interpretation based on the derivative? Just take the integral of the derivative, and you have the original formula... Right?"

OM: "Wait, what?" he reads the question above "Holy moly, I haven't done this in years!"

So, this lead to a discussion between us as to how to integrate this, but what the professor really wants (probably) is not to have you do the reverse operation (which in some cases may be really HARD), but to understand what the 1st derivative actually is.

So we scratched our heads and mulled through our collective age-addled memories, and finally agreed that the 2nd derivative is the local maxima/minima, and the 1st derivative (the one you care about) is the slope of the curve at the given point.

Well, what does this have to do with the price of worms in Mexico? Well, if we make an assumption that the slope remains relatively constant for all "nearby" points (to know this, you need to look at the curve and use good judgement based on what you know about things - but since this is what your prof wants, this is what he gets!), then we can do a linear interpolation - which is exactly what you asked for!

All right, then - the meat of the answer:
The slope (m) of the function at our known value is:

m=$\sqrt{{x}^{2} + 15}$

Therefore, the slope at out known point (x=1) is:

m=$\sqrt{{1}^{2} + 15}$
m=$\sqrt{1 + 15}$
m=$\sqrt{16}$
m=4

Remember, then, that the formula for a line (needed for linear interpolation) is:

$y = m x + b$

This means that for points "close" to our known value, we can approximate the values as being on a line with slope m, and y-intercept b. or:

$g \left(x\right) = m x + b$
$g \left(x\right) = 4 x + b$

So, then, what's $b$?

We solve for this using our known value:

$g \left(1\right) = 3$
$4 \left(1\right) + b = 3$
$4 + b = 3$
$b = - 1$

Now we know the formula for the line that approximates our curve at the known point:

g(x$\cong$1)=4x-1

So, no we insert our approximation points to get the approximate value, or:

$g \left(0.9\right) \cong 4 \left(0.9\right) - 1$
$g \left(0.9\right) \cong 3.6 - 1$
$g \left(0.9\right) \cong 2.6$

and

$g \left(1.1\right) \cong 4 \left(1.1\right) - 1$
$g \left(1.1\right) \cong 4.4 - 1$
$g \left(1.1\right) \cong 3.4$

Easy, right?