# Approximation and Logarithm Problem

#### phrox

##### New member
I just need some help with some basic questions I can't remember from a long time ago, just started up school again....

When at x=1, I am given a tangent line to the function f(x), and also given the equation of the tangent line. How can I use the tangent line to estimate f(any number). Also how to find error of approximation within each of these 3 points???

2) Solve for x: ln(7−2x)−4lnx=ln8 ( I changed values so I have to do work by myself after I understand this)

#### Ackbach

##### Indicium Physicus
Staff member
1) The equation of the tangent line is going to look like any line: $y=mx+b$. You can use the tangent line to approximate the function near the point of tangency, by simply evaluating the tangent line there.

2. Use some of the rules of logarithms to get a single logarithm on both sides. What do you get?

#### phrox

##### New member
Using tangent line to estimate, error of approx.?

I'm not going to post the question because I think that's defeating the purpose of me doing it on my own...

Say you have f(x) = (quadratic) / (quadratic), so (x^2 + x + 1) / ( x^2 + x + 1)...

I am given the tangent line at the point x = 1, how do I use this tangent line to estimate say f(0.5), f(0.6), f(0.65)? Also how can I find error of approximation at each of the 3 points? What is error of approximation?

#### MarkFL

Staff member
The example you give is a constant function, since the numerator and denominator are the same. Let us consider the function:

$$\displaystyle f(x)=\frac{2x^2-3x+1}{x^2+x+1}$$

The tangent line at $x=1$ is then:

$$\displaystyle y=\frac{x-1}{3}$$

A plot of the function and the tangent line is given here: As mentioned by Ackbach:

You can use the tangent line to approximate the function near the point of tangency, by simply evaluating the tangent line there.
The error $E(x)$ then is simply the magnitude of the difference between the approximation (the value of the tangent line for a particular $x$) and the actual value of the function, hence:

$$\displaystyle E(x)=\left|f(x)-y(x) \right|$$

#### phrox

##### New member
So am I supposed to plug the f(whatever) into the tangent line, which will give me the estimation of the number, then I just find the differences between all 3 intervals?

#### MarkFL

Staff member
To address the question (which has now been replaced) of how to find the derivative of a rational function consisting of a quadratic divided by a quadratic

One could use the quotient rule as follows:

Given:

$$\displaystyle f(x)=\frac{ax^2+bx+c}{dx^2+ex+f}$$

then:

$$\displaystyle f'(x)=\frac{\left(dx^2+ex+f \right)(2ax+b)-\left(ax^2+bx+c \right)(2dx+e)}{\left(dx^2+ex+f \right)^2}=\frac{(ae-bd)x^2+2(af-cd)x+(bf-ce)}{\left(dx^2+ex+f \right)^2}$$

For the replacement questions:

For $x$ near 1, we may use:

$$\displaystyle f(x)\approx y(x)$$

And the error is the magnitude of the difference between the approximation and the true value.

#### phrox

##### New member
Sorry, I edited that last question to a new one... Forgot I was given the tangent line hahah!

#### phrox

##### New member
Ok, I got it all except for the error of approximation...

I have f(x) = 0.25, f(x) = 0.85, f(x) = 0.925. This is all in my actual questions.

So if I plug in my given decimal numbers into the tangent line, I get those ^^^^^

If I plug in my same decimal numbers that I plugged into the tangent line into the very first eqn, I got -4 for the first one, etc etc. So this just means the E(x) = 4.25?