2007Fall1aMTI_sol

2007Fall1aMTI_sol - version1.1.3:47 Solutions to Midterm I...

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Unformatted text preview: version1.1.3,2008-10-1414:47 Solutions to Midterm I Math 1a Introduction to Calculus October 24, 2007 version1.1.3,2008-10-1414:47 1 Math 1a Solutions to Midterm I October 24, 2007 1 1. (10 Points) (a) (2 points) Let f be a function and a a point in the domain of f. State the definition of f ( a ) , the derivative of the function at a. Solution. The derivative of f at a , denoted f ( a ) , is the limit lim x → a f ( x )- f ( a ) x- a if it exists. This limit can also be written lim h → f ( a + h )- f ( a ) h The derivative does not exist for all functions f . N (b) (8 points) Let f ( x ) = x 2- 1 . Find an equation of the tangent line to the graph of f ( x ) at the point ( 1, 0 ) by using the definition of the derivative. No credit will be given for using the Power or any other Rules. Solution. The slope of the tangent line to the graph of f at a point ( a , f ( a )) is the derivative of f at a (if it exists). In this case, we have f ( a ) = lim h → f ( a + h )- f ( a ) h = lim h → ( a + h ) 2- 1- ( a 2- 1 ) h = lim h → a 2 + 2 ah + h 2- 1- a 2 + 1 h = lim h → 2 ah + h 2 h = lim h → ( 2 a + h ) Since 2 a + h is continuous as a function of h , we may plug in 0 to evaluate the limit and obtain f ( a ) = 2 a . In particular, f ( 1 ) = 2. The unique line passing through ( 1, 0 ) with slope 2 has the equation y = 2 ( x- 1 ) = 2 x- 2 N / 10 –1– version1.1.3,2008-10-1414:47 2 Math 1a Solutions to Midterm I October 24, 2007 2 2. (10 Points) Let f be given by f ( x ) = (- 2 x + 3 if x ≤ 1- x 2 + x + 1 if x > 1 Justify your answers to both of these: (i) Is f continuous at 1 ? Solution. lim x → 1- f ( x ) = lim x → 1 + (- 2 x + 3 ) = 1 lim x → 1 + f ( x ) = lim x → 1- (- x 2 + x + 1 ) = 1 f ( 1 ) = 1 so we have lim x → 1 f ( x ) = f ( 1 ) = 1 hence f ( x ) is continuous at 1. N (ii) Is f differentiable at 1 ? Solution. lim h → + f ( 1 + h )- f ( 1 ) h = lim h → +- 2 ( 1 + h ) + 3- ( 1 ) h =- 2 lim h →- f ( 1 + h )- f ( 1 ) h = lim h →-- ( 1 + h ) 2 + ( 1 + h ) + 1- ( 1 ) h = lim h →-- h + h 2 h =- 1 Since lim h → + f ( 1 + h )- f ( 1 ) h 6 = lim h →- f ( 1 + h )- f ( 1 ) h f ( 1 ) cannot exist. N For this test, we also accepted the following argument: for x < 1, f ( x ) =- 2. For x > 1, f ( x ) =- 2 x + 1. Since lim x → 1 + f ( x ) =- 1 and lim x → 1- 1 f ( x ) =- 2, f is not differ- entiable at 1. But the validity of this argument depends on the Mean Value Theorem, which we haven’t covered, so a better argument would only use the definition of the derivative, as above. / 10 –2– version1.1.3,2008-10-1414:47 3 Math 1a Solutions to Midterm I October 24, 2007 3 3. (10 Points) The population of a city is measured at two year intervals and is given as follows: Year 1984 1986 1988 1990 1992 1994 Population 265 290 324 358 395 437 Let P ( t ) be the population at any time t....
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2007Fall1aMTI_sol - version1.1.3:47 Solutions to Midterm I...

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