Calculus Test _2 solutions ver 1 - PC vers

Calculus Test_2 - Calculus Test#2 Solutions 1 So critical points occur at x=1 and x=0 This splits the function into 2 regions So Region I is

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Calculus Test #2 Solutions 1. So critical points occur at x=1 and x=0. This splits the function into 2 regions. So Region I is decreasing because f’(-1)=-24 Region II is decreasing because f’(0.5)=-1.5 Region III is increasing because f’(2)=48 2. f’’(x)=6x-12=0 x=2 Inflection point occurs at x=2. 3. f’(x)=2x-6=0 critical point is at x=3. YOU MUST COMPARE VALUE OF THE FUNCTION AT THE CRITICAL POINT WITH THE VALUE AT THE ENDPOINTS. f(0)=18 f(3)=9 f(5)=13 Therefore the global minimum is 9 and the global maximum is 18 4. One max, one min, THREE inflection points. 5. 1 st derivative is negative wherever the function is decreasing, positive when increasing, and 0 at critical points. 2 nd derivative is negative where the function is concave down, positive where concave up, and 0 at inflection points. 6. f’(x)=4x-6 f’(2)=2. So slope of tangent line is 2. Point 1 st Derviative 2 nd Derivative A 0 - B - 0 C 0 +
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F(2)=-4 so the point (2,-4) lies on the line Plug into y=mx+b -4=2(2)+b therefore b=-8
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This note was uploaded on 01/16/2012 for the course MATH 122 taught by Professor Kustin during the Spring '08 term at South Carolina.

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Calculus Test_2 - Calculus Test#2 Solutions 1 So critical points occur at x=1 and x=0 This splits the function into 2 regions So Region I is

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