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## Calculus 1

### Course: Calculus 1>Unit 5

Lesson 3: Intervals on which a function is increasing or decreasing

# Finding increasing interval given the derivative

Sal is given that the derivative of function g, is g'(x)=x²/(x-2)³. He uses that to find the intervals where g is increasing, by looking for the intervals where g' is positive.

## Want to join the conversation?

• In the previous video in the playlist 'CALCULUS DERIVATIVE APPLICATIONS' Sal put the derivative to be less than zero then just solved the inequality to find what x was in that case. Can't we just use that way for all problems of this kind? Which way is better? • Yes, you can solve it the way he did in the past video, he only wanted to show you that that wasn't the only way. In the example shown in this video, you would set both the numerator and the denominator greater to zero so your whole fraction can be greater than zero, by doing that you get that the x in the numerator should be greater than zero and the x in the denominator should be greater than 2. Since you're working with both at the same time, your x value should be greater than 2 so the whole fraction ends up greater than zero. Hope I helped!
• g'(x) is the slope of the function g,right? so when we get g'(x)=0,it would mean that the slope is parallel to x axis.but if the function has to go on after that point where sloe is 0,in this casewhen x=0,there should be a turning of the graph.
for eg,if the fn was decreasing frm (-infinity) till x=0,and at x=0,the slope is 0,so from x=0,shouldnt the graph increase??
here at the graph at both intervals are decreasing.how is that possible? • When g'(x)=0, x is a critical number (c.n.) A c.n. is where the graph could have min/max (turning point where the graph change from decrease to increase or vice versa) but it does not guarantee that it will have a min/max. That's why we have to do what we call the first derivative test like Sal does in the video.

An example of this would be f(x)=x³ then f'(x)=x²
f'(x) = 0 at x = 0, but f(x)=x³ is increasing for all x because at x=0 the slope is 0 but it's neither a min or a max.
• at , is there a difference between (negative infinity, 0) and [-negative infinity, 0]
is there a close interval contains negative infinity? • Parentheses ( indicate non inclusive. Brackets [ indicate inclusive. So (-inf,0) means it is neither negative infinity or zero, but everything in between. [-inf,0] means that it could be zero or negative infinity along with everything in between. Since something can never equal infinity, we use parentheses by them. therefore, [-inf,0] is technically impossible. (-inf,0) also does not equal (-inf,0] because it could be zero in the second one but not the first one.
• At the top of the screen, it says that g(x) is defined for all real numbers, but integrating g'(x) and plugging in x=2, we get that g(2) is undefined. In the example of the video, g'(2) is also undefined, but a critical value requires that it be defined by g(x). So x=2 isn't a critical value, and we can't define and interval using 2 as an endpoint. I understand that the results work out nicely with g(x) defined in the problem and the actual g(x) (which is undefined at x=2), but in general, without being given that some function f is continuous over all real numbers, wouldn't it be invalid to assume that f(x) has critical values where f'(x)=0 or undefined without knowing f(x) or integration? And furthermore, doesn't that mean it's invalid to define an interval using these values, and therefore you can't actually know where f(x) is increasing or decreasing based on f'(x) without knowing f(x) or integration? • What is the antiderivative of g'(x), i mean how do we integrate g'(x) = x^2/(x-2)^3 ? • It is not necessary to use partial fractions to integrate x^2/(x-2)^3.
We can use the u-substitution u=x-2. Then du = 1dx = dx, and x=u+2. Therefore, we have

integral of x^2/(x-2)^3 dx = integral of (u+2)^2 / u^3 du
= integral of (u^2+4u+4) / u^3 du
= integral of (1/u + 4/u^2 + 4/u^3) du
= ln|u| - 4/u - 2/(u^2) + C
= ln|x-2| - 4/(x-2) - 2/(x-2)^2 + C.

We can check this answer by differentiating it, and verifying that the result is the original function.
(d/dx) of (ln|x-2| - 4/(x-2) - 2/(x-2)^2 + C)
= 1/(x-2) + 4/(x-2)^2 + 4/(x-2)^3
= [(x-2)^2 + 4(x-2) + 4]/(x-2)^3
= (x^2 - 4x + 4 + 4x - 8 + 4)/(x-2)^3
= x^2/(x-2)^3.

The answer ln|x-2| - 4/(x-2) - 2/(x-2)^2 + C checks!

Have a blessed, wonderful day!
• feeling very dumb rn • In the graph of f(x)= polynomial having f'(x) = undefined could indicate that there's a sharp point in the graph of f(x) right where the slope would simultaneously be decreasing and increasing i.e. undefined? But what if you have f'(x) undefined in a function where f(x) = rational function? What would this undefined x in f'(x) indicate about the behavior as the graph approaches this x and what would that look like? • The point x=0 is also the inflection point, correct? • Why did we have to find the critical points–why didn't he just solve it like he did in the previous video? And what does this graph look like because I'm having trouble visualising it.  