2007Fall1aMTI_sol

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

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 15

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online