Simple Calculus Question, anyone?

Question

f(x) = x^2 - 2x + 1
g(x) = x^3 - 3x^2 + 2

find the inflection points
i know the answers but i dont know how to do the procedure
i remember you have to multiply exponents or something
can someone fully explain it to me, step by step ?

Answer 1

In order to find the inflection point, you must set the second derivative equal to zero and solve for x.

f(x)=x^2-2x+1

f '(x) = 2x-2

f ''(x)=2

2=0 (not true)

This one does not have an inflection point.


g(x) = x^3-3x^2+2

g '(x)= 3x^2-6x

g ''(x) = 6x-6

0=6x-6
6=6x
x=1

Therefore, there is an inflection point at x=1.

Answer 2

There are two conditions, the people above are only giving the first condition.

Take the second derivative and solve for zero (like they did). THEN, you have to test for a local min / local max, or an inflection point.

This test is done by taking the answer found in the second derivative, and solving for it +/- a very small value. If the signs are the same, then it is a local min or local max. If the signs are different, then it is an inflection point.

So, in the second example, the point to test is x=1, so you would test with x = 1.0000001 and x=.9999999

For the first test, the answer would be slightly positive, and in the second test it would be slightly negative.

Therefore it IS an inflection point (but simply setting the second derivative to zero is not sufficient to determine this.

Answer 3

An inflection point is a point on a curve at which the sign of the curvature (i.e., the concavity) changes.

Use the second derivative test to find an inflection point.

A necessary condition for x to be an inflection point is
f''(x) = 0.

A sufficient condition requires f''(x - ε) and f''(x + ε) to have opposite signs in the neighborhood of x.
__________

f(x) = x² - 2x + 1
f'(x) = 2x - 2 = 0
f''(x) = 2 ≠ 0 for all x.

There is no point of inflection.
____________

g(x) = x³ - 3x² + 2
g'(x) = 3x² - 6x = 0
g''(x) = 2x - 2 = 0
2x = 2
x = 1

g''(0.99) = -0.02 < 0
g''(1) = 0
g''(1.01) = 0.02 > 0

So the second derivative changes sign in the neighborhood of
x = 1. Therefore a point of inflection occurs at x = 1. The point of inflection is (1,0).
_____

If the sign of the second derivative had not changed in the neighborhood of x = 1, it would have been a relative maximum or minimum, not a point of inflection.

Answer 4

a necessary condition for inflection points is that the second derivative is zero at the inflection point.

differentiate these two functions twice (in other words, get the second derivative), set the second derivative equal to zero and find the value of x where the second derivative is zero

keep in mind that not every function has an inflection point