Re: Evaluating a function

There is nothing wrong. Both expressions give identical solutions for x.

Re: Evaluating a function

Thank you Gusbob I see that you are correct :)

Re: Evaluating a function

If I now continue to find the critical numbers (using both method one and two):

**Method one:**

$\displaystyle 0=(2x^2-x-1)/x$

$\displaystyle 0=((2x+1)(x-1))/x$

The critical numbers should be:

$\displaystyle x=-1/2$

$\displaystyle x=1$

and $\displaystyle x=0$ (from the denominator)

**?**

**Method two:**

$\displaystyle 0=(2x^2-x-1)$

$\displaystyle 0=((2x+1)(x-1))$

The critical numbers should be:

$\displaystyle x=-1/2$

$\displaystyle x=1$

**?**

My final question is.. why is there **3 critical numbers using method one** but only **2 critical numbers using method two**? Am I thinking worng?

Re: Evaluating a function

Only the numerator will make the quotient equate to zero. Division by zero is undefined so you can neglect the denominator.

So $\displaystyle \frac{2x^2 - x -1}{x} = 0 $ reduces to $\displaystyle (2x + 1)(x - 1) = 0 $

Re: Evaluating a function

No, you cannot "neglect the denominator" for **this** problem. astarledeer is correct that a fraction, and so the dervative, is 0 only when the numerator is 0, independent of the denominator. But this problem asked for "critical points" which are points at which the derivative is 0 **or** where the derivative does not exist. Going back to the original equation,

f'= 2x- 1- 1/x, f' does not exist at x= 0. However, now we note that the original function, f(x)= x^2- x- ln(x) is also not defined at x= 0. Therefore, the only critical points ** in the domain of the original function** are x= 1/2 and x= 1. But there are examples where the function exists and some point but the derivative does not: $\displaystyle y= \sqrt{x}$ is one.

Re: Evaluating a function

Duffman

Mathematics is not just doing additions or subtractions and ..other operations but rather thinking and reasoning.

first of all do you know what a critical point is?

if not go here to revise.... Critical point (mathematics) - Wikipedia, the free encyclopedia

in your case the critical points are 3 as you said :point zero where the derivative does not exist.....this is obvious from the domain of your function...and points 1 and (-1/2)??? but you miss the point again..... the value x= -1/2 is out of the domain of your function...in other words your function is defined in the domain (0,+infinity)..

and x = -1/2 is not included....therefore the only available critical point to consider is the point x=1.

MINOAS

Re: Evaluating a function

Sorry Hall

your post was not there when I started mine.