When you're finding the sign of f' and f'' to find a local maximum/minimum, concavity and inflection point where do you substitute the tangent/critical points? In which function? f(x), f'(x) or f''(x)?
2 answers
First determine the points where f'(x) = 0 Then evaluate f''(x) at those same points to determine inflection, concavity and maximum/minimum.
rather than going through a lengthy explanation, I will use an example to answer your question.
e.g. f(x) = x^3 - 6x^2 + 1
f'(x) = 3x^2 - 12x
set this to zero, to find local max/mins
3x^2 - 12x = 0
3x(x-4) = 0
x = 0 or x = 4, then
y = 1 or y = -31
then (0,1) and (4,-31) are local max/mins, but I don't know at this point whether they are a maximim or minimum
f''(x) = 6x-12
if f''(x) is positive for some x, then the curve is concave upwards at that point, and if f''(x) is negative for some x, then the curve is concave downwards at that x.
so we sub (0,1) into 6x-12 to get -12
so at (0,1) the curve opens downwards, so (0,1) must be a maximum point
similarly at (4,-31), f''(4) = 24-12 = 12, positive, so at (4,-31) the point opens upwards, implying (4,-31) is a minimum point.
notice when I set f''(x) = 0,
6x-12 = 0 I get x = 2, and y = -15
so the point of inflection is (2,-15)
which happens to be the midpoint between the max and min points. This is always true for a general cubic function, and makes an interesting property to prove.
Also notice that f'(x) indicates whether the curve is increasing or decreasing at any x value that you pick.
Make yourself a summary to learn which derivative is used for what.
e.g. f(x) = x^3 - 6x^2 + 1
f'(x) = 3x^2 - 12x
set this to zero, to find local max/mins
3x^2 - 12x = 0
3x(x-4) = 0
x = 0 or x = 4, then
y = 1 or y = -31
then (0,1) and (4,-31) are local max/mins, but I don't know at this point whether they are a maximim or minimum
f''(x) = 6x-12
if f''(x) is positive for some x, then the curve is concave upwards at that point, and if f''(x) is negative for some x, then the curve is concave downwards at that x.
so we sub (0,1) into 6x-12 to get -12
so at (0,1) the curve opens downwards, so (0,1) must be a maximum point
similarly at (4,-31), f''(4) = 24-12 = 12, positive, so at (4,-31) the point opens upwards, implying (4,-31) is a minimum point.
notice when I set f''(x) = 0,
6x-12 = 0 I get x = 2, and y = -15
so the point of inflection is (2,-15)
which happens to be the midpoint between the max and min points. This is always true for a general cubic function, and makes an interesting property to prove.
Also notice that f'(x) indicates whether the curve is increasing or decreasing at any x value that you pick.
Make yourself a summary to learn which derivative is used for what.