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## Calculus 1

### Unit 5: Lesson 10

Connecting f, f', and f''- Calculus-based justification for function increasing
- Justification using first derivative
- Justification using first derivative
- Justification using first derivative
- Inflection points from graphs of function & derivatives
- Justification using second derivative: inflection point
- Justification using second derivative: maximum point
- Justification using second derivative
- Justification using second derivative
- Connecting f, f', and f'' graphically
- Connecting f, f', and f'' graphically (another example)
- Connecting f, f', and f'' graphically

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# Calculus-based justification for function increasing

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.10 (EK)

, FUN‑4.A.11 (EK)

A calculus-based justification is when we explain a property of a function f based on its derivative f'. See a good example (and a few wrong ones) for how to do this when explaining why a function increases.

## Want to join the conversation?

- Can someone confirm if this is correct? I am trying to learn this stuff for my calc test and want to make sure I understand it

1) f(x) is concave up —> f'(x) is increasing —> f''(x) graph values are positive

2) An inflection point on f(x) corresponds to a max/min on f'(x) and a x-intercept on f''(x)(4 votes)- Let's start with your first statement:

1.) Yes, if f(x) is assumed concave up, f'(x) must be increasing on the concaved up interval, and therefore, f''(x) must be positive on this same interval.

-If f'(x) is increasing, it could still be negative until it would pass a critical point (f'(x) = 0) and then f'(x) would turn positive.

-The 2nd derivative, f''(x) being positive is implying a positive rate of change of the first derivative, f'(x). So, this would imply the original function, f(x) is increasing more greatly in a positive, upward direction, thus, the concaved upward shape.

-This circumstance you have stated would imply a local or relative minimum of the function, f(x). In other words, if f'(c) = 0 and f''(c) > 0, then there is a local minimum at the point x=c.

2.) No, an inflection point may not necessarily mean a maximum/minimum on f'(x). For example, the max/min of f'(x) at some point x=c is stating that f''(c) = 0. This doesn't mean it's an inflection point because we do not know if f''(x) is changing signs from negative to positive or positive to negative. The derivative f'(x) could be increasing, become zero, and then continue to increase afterward. This would state that f''(x) is always positive--the slope of the first derivative is increasing on that interval. The definition of an inflection point is just a point where the concavity changes signs, so f''(x) > 0 would switch to f''(x) < 0 and vice-versa. Looking at a point, x=c, if f''(c-0.0001) < 0 and f''(c+0.0001) > 0, this would imply an inflection point at the point c.

-As for the inflection point being an x-intercept on f''(x), this could be true, but it is not necessarily true. It's an 'x-intercept' in the sense that you're setting f''(x) = 0 and solving for the x-values; however, these are just candidate inflection points. You will have to test the interval before and after the said candidate inflection point to determine whether it is or not. If f''(x) changes signs, then yes, it is an inflection point.(2 votes)

- In the video, Sal said that an increasing/decreasing derivative doesn't mean that the function is increasing/decreasing--- Yet is there an alternative meaning for that ?(2 votes)
- There is a difference between increasing and positive. if the derivative is positive at a point, then the original function must be increasing at that point. However, the derivative can be increasing without being positive. For example, the derivative of f(x) = x^2 is 2x. if you graph f'(x) = 2x, you can see that for any negative x value, the graph is negative. However, f'(x) is still increasing; it is becoming less negative. So in this case, the derivative is increasing, but the function is decreasing.(3 votes)

- Hi, why is the answer "The derivative of h is positive when x > 0" correct? The derivative of a function may be decreasing even when it is positive where x > 0 such as the part of the curve of a function which is concaving downwards before reaching the max point.

Thanks.(2 votes)- You are correct in saying that the statement is incorrect in the general sense. However in the specific case of the graph given, it is true, which is what the statement is referring to.(1 vote)

- Should we expect 2020 AP Calculus AB test to have questions like this? It can't be googled easily so I suppose its a good question style to avoid cheating at home.(0 votes)

## Video transcript

- [Instructor] We are told
the differentiable function h and its derivative h prime are graphed and you can see it here, h is in blue and then its derivative h
prime is in this orange color. Four students were asked to give an appropriate
calculus-based justification for the fact that h is increasing when x is greater than zero. Can you match the teacher's
comments to the justifications? So before I even look at
what the students wrote, you might say, hey, look,
I can just look at this and see that h is increasing
when x is greater than zero but just by looking at the graph of h, that by itself is not a
calculus-based justification. We're not using calculus. We're just using our knowledge of what it means for a
graph to be increasing. In order for it to be a
calculus-based justification, we should use calculus in some way. So maybe use the derivative in some way. Now, you might recognize that you know that a
function is increasing if its derivative is positive. So before I even look at
what the students said, what I would say, my
calculus-based justification, and I wouldn't even have
to see the graph of h, I would just have to
see the graph of h prime is to say, look, h prime is greater than, h prime is positive when
x is greater than zero. If the derivative is positive then that means that the slope of the tangent line is positive and that means that the graph
of the original function is going to be increasing. Now, let's see whether one
of the students said that or what some of the other students wrote. So can you match the teacher's comments to the justifications? So one student wrote, the
derivative of h is increasing when x is greater than zero. So it is indeed the case that
the derivative is increasing when x is greater than zero but that's not the justification
for why h is increasing. For example, the derivative
could be increasing while still being negative in which case h would be decreasing. The appropriate justification
is that h prime is positive, not that it's necessarily increasing 'cause you could be increasing
and still not be positive. So let's see. I would say that this doesn't
justify why h is increasing. When x is greater than zero, as the x-values increase, the
function values also increase. Well, that is a justification
for why h is increasing but that's not calculus-based. In no way are you using a derivative. So this isn't a
calculus-based justification. It's above the x-axis. So this one, what is it? Are they talking about h? Are they talking about h prime? If they were saying that h
prime is above the x-axis when x is greater than zero
then that would be a good answer but this is just what is above the x-axis and over what interval? So I would actually, let's
scroll down a little bit, this looks like a good thing
for the teacher to write. Please use more precise language. This cannot be accepted as
a correct justification. And then finally, this last student wrote, the derivative of h is positive
when x is greater than zero and it is indeed the case. If your derivative is positive, that means that your original function is going to be increasing
over that interval. So kudos, you are correct.