Critical Points and Concavity Analysis of a Function - Prof. Michael Rosenthal, Study notes of Calculus

An analysis of a function's critical points and concavity. The function is represented by the given equation, and the critical points are identified at x = 2 and x = 4. The function is decreasing for x < 2 and increasing for x > 4. The function has a relative maximum at x = 4 and a relative minimum at x = 2. The function is concave up for x < 3 and concave down for x > 3. It also has an inflection point at (3, 0).

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Pre 2010

Uploaded on 09/17/2009

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18249)( 23 +โˆ’+โˆ’= xxxxf
)4)(2(3)86(324183)( 22 โˆ’
โˆ’
โˆ’
=
+
โˆ’
โˆ’
=โˆ’+โˆ’=
โ€ฒxxxxxxxf
So we have critical points at x = 2, 4
- 3 - - - - - - - - - - - - - - - - - - - - - - - - -
x โ€“ 2 - - - -0 + + + + + + + + + + + + + + +
x โ€“ 4 - - - - - - - - - - - - - - - - 0 + + + + + +
f is decreasing on ),4()2,(
โˆž
+
โˆ’
โˆžand
f is increasing on (2, 4)
f has a relative maximum of 2 at x = 4
f has a relative minimum of -2 at x = 2
)3(6186)(
โˆ’
โˆ’=+โˆ’=
โ€ฒโ€ฒ xxx
f
-6 - - - - - - - - - - - - - - - - - - - - - - - - -
x โ€“ 3- - - - - - - - - - - - - 0 + + + + + + + +
f is concave up on )3,(โˆ’โˆž
f is concave down on ),3( โˆž+
f has an inflection point of (3, 0)
24
-2
2
x
y
3
2
4

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3 2 f x = โˆ’ x + x โˆ’ x + ( ) 3 18 24 3 ( 6 8 ) 3 ( 2 )( 4 ) 2 2 f โ€ฒ^ x =โˆ’ x + x โˆ’ =โˆ’ x โˆ’ x + =โˆ’ x โˆ’ x โˆ’ So we have critical points at x = 2, 4

  • 3 - - - - - - - - - - - - - - - - - - - - - - - - - x โ€“ 2 - - - -0 + + + + + + + + + + + + + + + x โ€“ 4 - - - - - - - - - - - - - - - - 0 + + + + + + f is decreasing on ( โˆ’โˆž, 2 ) and ( 4 ,+โˆž) f is increasing on (2, 4) f has a relative maximum of 2 at x = 4 f has a relative minimum of -2 at x = 2 f โ€ฒโ€ฒ^ ( x )=โˆ’ 6 x + 18 =โˆ’ 6 ( x โˆ’ 3 ) -6 - - - - - - - - - - - - - - - - - - - - - - - - - x โ€“ 3- - - - - - - - - - - - - 0 + + + + + + + + f is concave up on (โˆ’โˆž, 3 ) f is concave down on ( 3 ,+โˆž) f has an inflection point of (3, 0) 2 4

2 x y 3 2 4