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Sample-free-response-midterm-solutions

# Sample-free-response-midterm-solutions - 2 x 1 x < 2 1...

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1. Suppose 2 1, 2 ( ) 3, 2 5 2, 5 x x f x x x x + < - = - - Use the three-part test for continuity at a point to determine if the function is continuous at x = -2 and at x = 5. If the function is not continuous at one (or both) of the points, be sure you state which one of the three parts is not satisfied. State the condition itself (i.e., not I, II or III). First, at x = -2. 2 2 2 2 2 ( 2) 3 lim ( ) 2( 2) 1 3 lim ( ) 3 So lim ( ) does not exist, since lim ( ) lim ( ) x x x x x f f x f x f x f x f x - + - + →- →- →- →- →- - = = - + = - = Since the limit doesn’t exist, f is not continuous at x = -2. Second, at x = 5. 5 5 5 5 5 5 (5) 3 lim ( ) 3 lim ( ) 5 2 3 So lim ( ) exists, since lim ( ) lim ( ). Finally, (5) lim ( ) 3. x x x x x x f f x f x f x f x f x f f x - + - + = = = - = = = = We can conclude that the function is continuous at x = 5.

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2. Use the limit definition of the derivative (four-step process) to find the derivative: 2 6 3 ) ( 2 + - = x x x f 2 2 2 2 2 2 2 2 2 2 2 2 2 0 ( ) 3( ) 6( ) 2 3( 2 ) 6( ) 2 3 6 3 6 6 2 ( ) ( ) 3 6 3 6 6 2 (3 6 2) 3 6 3 6 6 2 3 6 2 6 3 6 ( ) ( ) 6 3 6 (6 3 6) 6 3 6 ( lim h f x h x h x h x xh h x h x xh h x h f x h f x x xh h x h x x x xh h x h x x xh h h f x h f x xh h h h h h x h h x h f x + = + - + + = + + - + + = + + - - + - = + + - - + - - + = + + - - + - + - = + - + - + - = + - = = + - 0 ) ( ) lim(6 3 6) 6 6 h h f x x h h x + - = + - = - 3. Find the first derivative: 4 3 ) ( 2 2 - = x x x f We’ll need to use the quotient rule. ( 29( 29 ( 29( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 2 2 2 2 2 2 2 3 3 2 2 2 2 4 3 3 4 ( ) 4 4 2 3 (2 ) 4 2 8 2 6 4 14 4 x x x x f x x x x x x x x x x x x x x + - - - + = + + - - = + + - + = + = +
4. Find the first derivative: x x x f 3 ) ( 2 - = We’ll need to use the chain rule. ( 29 ( 29 ( 29 ( 29 1 2 2 1 2 2 1 2 2 2 ( ) 3 1 ( ) 3 2 3 2 2 3 2 3 2 3 2 3 f x x x f x x x x x x x x x x - = - = - - - = - - = - 5. Find the first and second derivatives: ( 5 2 5 2 ) ( + = x x f We’ll need to use the chain rule. ( ( 29 ( 29 5 2 4 2 4 2 ( ) 2 5 ( ) 5 2 5 (4 ) 20 2 5 f x x f x x x x x = + = + = + To find the second derivative, we’ll need to use both the product and chain rules. ( ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 4 2 4 4 2 2 4 2 3 2 3 4 2 2 2 ( ) 20 2 5 ( ) 20 2 5 2 5 (20 ) (20 ) 4(2 5) (4 ) 2 5 20 320 2 5 20 2 5 f x x x f x x x x x x x x x x x x = + ′′ = + + + = + + + = + + +

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6. Find the first derivative: ( 4 3 2 3 2 ) ( + = x x x f We’ll need the product rule and the chain rule.
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