calculus question?

Question

How do you use the first and second derivitive to determine max and mins, inflection points, concavity, etc on a graph?

Answer 1

If f(x) is continuous and differentiable over x = a,b, then f'(x) = 0 indicates a min or max. Why? Because f'(x) = 0 means the slope of the curve is either horizontal at the top of the curve (a max) or at the bottom (a min). Solve f'(x) = 0 to find x = d, the X value where the max or min occurs. But which is it...a max or a min...read on.

If f''(x = d)>0, the point where f'(x = d) = 0 is a min because f''(x = d)>0 indicates the curve is changing direction upward (positive)...that is, it's bottomed out and coming back up. Naturally enough, if f''(x = d)<0, f'(d) would indicate a max because the curve is turning downward (negative).

When f''(x = d)>0, we say the curve is concave upwards (that happens at the min point x = d). The curve is concave downwards where f'(x = d)<0 at the max point d.

An inflection point occurs where the second derivative (f''(x)) changes sign (plus to minus or vice versa). Clearly x = d where the min or max occurs, is one such point because a rising curve starts down or a falling curve starts up at these special points of inflection.

Other inflection points might exist as well. Determine the second derivative and then identify where on x its value changes sign. That's an inflection point. Make sure you find all the points, there may be many.

Answer 2

equate the first derivative formula to zero and solve for x. if x is a min, the second derivative is greater than zero; if x is a max, the second derivative is smaller than zero.

alternatively, you can do a first derivative test on numbers that are smaller and greater than the min or max to find the slopes on both sides of the x value.

for example, the formula y=x^2 has the first derivative 2x. equate 2x = 0 and you get x = 0. second derivative = 2, meaning that x=0 is a min point. if u use the first derivative test, you find that substituting x= -1 into 2x gives -2 (negative gradient/slope) and substituting x = 1 gives 2 (positive gradient/slope). you can then deduce that x=0 is a min pt.

the first derivative test must be used when the second derivative gives 0 and there is no guarantee what x is ( min, max, or pt of inflexion ) when second derivative is 0.

to find concavity, if second derivative is greater than 0, the graph concaves upwards like a u shape. if the second derivative is smaller than 0, the graph concaves downwards like an inverted u.

final advice, if in doubt, try to plot the graph when possible. graphmatica is a good software for that

Answer 3

the derivivative is the slope of the curve at a point. if you set your first deriviative to 0, then solve for x, you will find locations of local minima and maxima.

Inflection point is where the 2nd derivative is zero -> same deal. fidn the 2nd derivative, set it to 0, solve for x. I can't remmber right off the top of my head about convacity -- but I think it may have to do with the second derivative being positive... but don't quote me on that.

Answer 4

Yes.. If the second derivative is positive the function is convex, otherwise it's concave.

Answer 5

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