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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)=6x12=0
x=2
Inflection point occurs at x=2.
3.
f’(x)=2x6=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)=4x6 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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View Full Document 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.
 Spring '08
 KUSTIN
 Calculus, Critical Point

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