Question 1105402
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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 <font color=blue>negative</font>


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 <font color=red>positive</font>


The second derivative goes from <font color=blue>negative</font> to <font color=red>positive</font> as it passes through x = 1
This means the concavity of f(x) changes from <font color=blue>concave down</font> to <font color=red>concave up</font>
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 <font size=5>(1,4)</font>
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