SOLUTION: The inflections point for f(x)=x^3-3x^2 +x+5 is

Question 1105402: The inflections point for f(x)=x^3-3x^2 +x+5 is
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

f(x)=x^3-3x^2+x+5 .... original function
f ' (x)=3x^2-6x+1 ... first derivative
f '' (x)=6x-6 ... second derivative

Solving f '' (x) = 0 leads to
6x-6 = 0
6x = 6
x = 1

The possible inflection point is at x = 1. We need to see if the second derivative changes sign as it passes through x = 1
Plug in an x value to the left of x = 1. Say x = 0
f '' (x)=6x-6
f '' (0)=6(0)-6
f '' (0)=-6 ... note how the result is negative

Now plug in a value to the right of x = 1. Say x = 2
f '' (x)=6x-6
f '' (2)=6(2)-6
f '' (2)=6 ... note how the result is positive

The second derivative goes from negative to positive as it passes through x = 1
This means the concavity of f(x) changes from concave down to concave up
Due to the concavity change, this means the inflection point for f(x) is at x = 1.

To find the y coordinate of the inflection point, plug x = 1 back into the original f(x) function
f(x)=x^3-3x^2+x+5
f(1)=(1)^3-3(1)^2+(1)+5
f(1)=1-3(1)+(1)+5
f(1)=1-3+1+5
f(1)=-2+1+5
f(1)=-1+5
f(1)=4

Answer: the inflection point for f(x) is at (1,4)

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