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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.

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• 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)

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.
• 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 ?
• 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.
• 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.