Stupid questions threat

why the fuck does the second derivative give the inflection point, I don't understand.

also, stupid question threat

its the slope of the first derivative?

I guess I haven't really answered it but another interesting way to think about it would be;

Why is the derivative of a distance-time graph equivalent to a velocity-time graph? Same for v(t) to a(t) [acceleration]

I can understand derivatives in terms of velocity / acceleration
But i'm having a hard time picturing (or understanding) how inflection points are given when setting the second derivative equal to 0

because the derivative of a generic differentiable function f(x) is the slopes/tangents of f(x)

visualising this: where f(x) has a maxima/minima the slope is zero so f'(x) will be zero at that point

where f(x) is increasing/decreasing the greatest it will have a peak for f'(x)

continuing this logic you can see that when you take the derivative of f'(x) the peaks become zero's on f''(x) which tell you where the slopes of f(x) are the greatest and the neighbouring points have positive or negative values depicting the concavity

can you write the third point in another way; having trouble understanding

the continuation logic or the peaks for f'(x)?

If you can understand how the zeroes of the first derivative give mins and maxes, recognize that the zeroes of the second derivative give mins and maxes of the first derivative, which correspond to inflection points of the original function for a very similar reason -- inflection points are where the slope changes from increasing to decreasing or vice versa.

I understand how the derivative of anything will tell you max and mins, but I wanna think in terms of f(x), not f'(x)
I'm just having trouble visualizing it all.

if anyone can do an example with f(x) = x^4

I was suggesting you trace back to f(x) in two steps instead of directly. f'(x) itself describes f(x), so using it as an intermediate can be helpful.

f(x) = x^4 does not have an inflection point, f(x) = x^3 has a rather trivial one at x=0.

you're going to have to think about f'(x) if you want to understand the second derivative of f(x)

f(x) = x^4
f'(x) = 4x^3
f''(x) = 12x^2

since f''(x) α (proportional to) x^2 it's positive at all values of x therefore f(x) is concave up at all values of x

think about what the peaks of f'(x) are compared to f(x) and then apply the logic of peaks of f(x) to f'(x)

I'll try to understand this tomorrow, but thanks for trying to help.

How do you find the maximum rectangle inside the area between sin(x) and the x axis?
Obviously it will have two points in the x-axis and the other two will be in the sin(x) function. Also the height of such a rectangle is sin(c) and the length pi-2c for a certain c. But I don't know where to go from there.

> derivatives
you don't deserve shit

Write an equation representing the area of such a triangle, sin(x)*(pi-2x), find zeroes of the derivative in the appropriate range and check which is the maximum. If this isn't for homework, I'd switch to cos and divide the problem in half via symmetry, working from 0 to pi/2. I always try to simplify ahead of time, but it shouldn't be much different either way.

How can I make Emacs work like TeXmaker?
I.e. tab-completion, match parenthesis etc.
Can't get this shit to work however hard I try. Any ideas?

just think of f''(x) as the acceleration of f(x)

if f''(x) is positive f(x) is accelerating up so it's concave up

Best online resource to learn calculus?

hello? anyone? I'm sure someone learned calc online...

I'm going to assume you mean calc without proofs, get a textbook like stewarts calculus since it has a fuckload of practice problems and supplement with patrickJMT. He does pretty good explanations. Pauls Online Math Notes are good too.

>he doesnt understand derivatives

are you 12?

Define inflection point without using derivatives, then your question will make sense.

that actually makes sense, thanks

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