Concavity and the 2nd Derivative Test
New Vocabulary:
- Inflection Point:
- Point where the concavity of the graph changes sign.
- Found where f''=0 or undefined
- Concavity
- The curve or shape of a line
- f''>0 the function is concave up (smily face)
- f''<0 the function is concave down (frowny face)
Example:
Determine the open interval on which f(x) is concave up and/or concave down, and all inflection points. f(x)=6 / (x^2+3)
ONE: Find first and second derivative
f(x)=6/(x^2+3) **Rearrange function so it is easier to take the first derivative
=6(x^2+3)^-1
f'(x)= -6 (x^2+3)^-2 (2x) **Take first derivative and simplify
= -12x / (x^2+3)^2
f''(x)= [(x^2+3)^2 (-12) - (-12x) (2) (x^2+3) (2x)] / (x^2+3)^4 **Take the derivative
= (x^2+3) (-12(x^2+3) + 48x^2) / (x^2+3)^4 **Factor out (x^2+3) then cancel with bottom
= -12 [(x^2+3) - 4x^2] / (x^2 +3)^3 ** Factor out -12
= -12 (-3x^2+3) / (x^2 +3)^3 ** Simply top
= 36 (x^2-1) / (x^2+3)^3 ** Factor out -3 and multiply with -12
TWO: f''(x) = 0 or undefined
f''(x)= 36 (x^2-1) / (x^2+3)^3
36 (x^2-1) = 0 ** Set top equal to zero to find POSSIBLE inflection points
x= -1, 1 ** POSSIBLE Inflection points
Bottom is never undefined! (x^2 +3)^3 can never equal zero
THREE: Plug inflection points into original to find the y-values to the point (use calc.)
f(1) = 6 / (1^2 + 3)^2 = 3/2
f(-1) = 6 / (-1^2 +3)^2 = 3/2
FOUR: Create a number line to test the inflection points
FIVE: Answer the question
Concave up: (1, infinity) and (- infinity, -1)
Concave down: (-1,1)
Inflection Points: (1, 3/2) and (-1, 3/2)
ONE: Find first and second derivative
f(x)=6/(x^2+3) **Rearrange function so it is easier to take the first derivative
=6(x^2+3)^-1
f'(x)= -6 (x^2+3)^-2 (2x) **Take first derivative and simplify
= -12x / (x^2+3)^2
f''(x)= [(x^2+3)^2 (-12) - (-12x) (2) (x^2+3) (2x)] / (x^2+3)^4 **Take the derivative
= (x^2+3) (-12(x^2+3) + 48x^2) / (x^2+3)^4 **Factor out (x^2+3) then cancel with bottom
= -12 [(x^2+3) - 4x^2] / (x^2 +3)^3 ** Factor out -12
= -12 (-3x^2+3) / (x^2 +3)^3 ** Simply top
= 36 (x^2-1) / (x^2+3)^3 ** Factor out -3 and multiply with -12
TWO: f''(x) = 0 or undefined
f''(x)= 36 (x^2-1) / (x^2+3)^3
36 (x^2-1) = 0 ** Set top equal to zero to find POSSIBLE inflection points
x= -1, 1 ** POSSIBLE Inflection points
Bottom is never undefined! (x^2 +3)^3 can never equal zero
THREE: Plug inflection points into original to find the y-values to the point (use calc.)
f(1) = 6 / (1^2 + 3)^2 = 3/2
f(-1) = 6 / (-1^2 +3)^2 = 3/2
FOUR: Create a number line to test the inflection points
- Test x= -2, 0, 1 in f''(x) to see if positive or negative (shown below test number)
- Where f''(x)>0 the function is concave up
- Where f''(x)<0 the function is concave down
FIVE: Answer the question
Concave up: (1, infinity) and (- infinity, -1)
Concave down: (-1,1)
Inflection Points: (1, 3/2) and (-1, 3/2)
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