Calculus Midterm Exam October 31, 2018

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1 Calculus Midterm Exam October 31, Use l Hôpital s Rule to evaluate the it, if it exists. tan3x (a) (6 points) sinx tan3x = 0, sinx = 0, and both tan3x and sinx are differentiable near x = 0, tan3x sinx = 3sec 3x cosx = 3 by l Hôpital s Rule. (b) (6 points) ( cscx cotx ) cscx =, cotx =, the it is of the indeterminate form. ( ) 1 cosx sinx cscx cotx = = = 0 by l Hôpital s Rule. sinx cosx. (6 points) Use the Squeeze Theorem to prove that x cos x = 0. Solution: For any x 0, since 0 x cos x x and we have 0 = 0 = x, x cos x = 0 by the Squeeze Theorem. x cos x = (8 points) Let f be the Dirichlet function f (x) = Prove that f is discontinuous at any x R. 0 if x is rational 1 if x is irrational. Solution: Take ε = 1. For any x R = Q ( R \ Q ) and for any δ > 0, there exist r Q (x δ,x + δ) and z ( R \ Q ) (x δ,x + δ) such that either if x Q = f (x) f (z) = 0 1 = 1 > 1 = ε

2 Calculus Midterm Exam October 31, 018 or if x R \ Q = f (x) f (r) = 1 0 = 1 > 1 = ε. Either case proves that f is discontinuous at x. Since x is arbitrary, f is discontinuous on R. 4. Suppose that Find the followin values. (a) (4 points) ( f ) (5) f (5) = 1, f (5) = 6, (5) = 3, and (5) =. Solution: If both f and are differentiable at x then, by the Product Rule, ( f ) (x) = f (x)(x) + f (x) (x). Therefore ( f ) (5) = f (5)(5) + f (5) (5) = 6 ( 3) + 1 = 16. (b) (4 points) ( f ) (5) Solution: If both f and are differentiable at x and (x) 0 then, by the Quotient Rule, ( f ) (x) f (x)(x) f (x) (x) =. (x) Therefore ( f ) (5) f (5)(5) f (5) (5) = = 6 ( 3) 1 (5) ( 3) = (a) (6 points) Find the derivative of y = cos 4 ( sin 3 x ). Solution: By the Chain Rule, dy dx = 4 cos 3 ( sin 3 x ) ( 1) sin ( sin 3 x ) 3 sin x cosx = 1 cos 3 ( sin 3 x ) sin ( sin 3 x ) sin x cosx (b) (6 points) Find d35 dx 35 ( xsinx ). Solution: For j and k 1, since d j x dx j = 0 and d4 j sinx dx 4 j = sinx, and by the binomial formula and the Product Rule, we have d 35 ( ) dx xsinx = 35 dx 35 dx d34 sinx dx 34 +x d35 sinx dx 35 = 35 d sinx dx +x d3 sinx dx 3 = 35 sinx x cosx. (c) (6 points) Use implicit differentiation to find an equation of the tanent line to the curve x /3 + y /3 = 4 at the point ( 3 3,1). Pae

3 Calculus Midterm Exam October 31, 018 Solution: By usin implicit differentiation, we have ( x 1/3 + y 3 Thus, at the point ( 3 3,1), we have 1/3 dy dx ) dy = 0 = dx = ( y) 1/3. x dy dx = ( 1 3 ) 1/3 ( 1 ) 1/3 1 = = /. the tanent line to the curve x /3 + y /3 = 4 at the point ( 3 3,1) has an equation y 1 = 1 3 (x + 3 3). 6. (8 points) Find the absolute maximum and absolute minimum values of f (x) = 3x 4 4x 3 1x + 1 on the interval [,3]. f (x) = 1x 3 1x 4x = 1x(x x ) = 1x(x + 1)(x ), the citical numbers of f in [,3] are x = 1, or 0, or. At the critical numbers, we have f ( 1) = = 4, f (0) = 1, f () = = 31. At the endpoints, we have f ( ) = = 33, f (3) = = 8. max f (x) = f ( ) = 33 and min f (x) = f () = 31. x [,3] x [,3] 7. (a) (6 points) Let f be continuous at c and assume that f (c) > 0. Prove that there exists δ > 0 such that f (x) > 0 for all x (c δ,c + δ). f (c) > 0 = f (c) > 0. By takin ε = f (c), the ε δ definition of the continuity of f at c implies that there exists a δ > 0, such that if x (c δ,c + δ) then f (x) f (c) < f (c) = f (c) < f (x) f (c) = f (x) > f (c) > 0. (b) (6 points) Let f be differentiable at c. Prove that f is continuous at c. Pae 3

4 Calculus Midterm Exam October 31, 018 f be differentiable at c, and f (x) f (c) = x c x c f (x) f (c) x c f (c) = x c f (x) f (c) x c exists, (x c) = f (x) f (c) x c x c (x c) x c = f (c) 0 = 0. and f is continuous at c. f (x) = f (c) x c 8. Let f be differentiable on (a, b). (a) (6 points) Suppose that f (x) 0 for all x (a,b). Prove that f is 1 1 on (a,b). Solution: For any x 1 x (a,b), there exits c lyin between x 1 and x, by the Mean Value Theorem, such that f (x 1 ) f (x ) = f (c)(x 1 x ) 0 = f (x 1 ) f (x ) = f is 1 1 on (a,b). (b) (6 points) Suppose that f (x) > 0 for all x (a,b). Prove that f is increasin on (a,b). Solution: For any x 1 < x (a,b), there exits c (x 1,x ), by the Mean Value Theorem, such that f (x 1 ) f (x ) = f (c)(x 1 x ) < 0 = f (x 1 ) < f (x ) = f is increasin on (a,b). 9. (16 points) For x ±1, let f (x) = x x 1. (a) Find the intervals on which the raph of y = f is increasin or decreasin, critical numbers and local extreme values, if exist. f (x) = (x 1)(4x) x x (x 1) = 4x (x 1) > 0 if x (, 1) ( 1,0), < 0 if x (0,1) (1, ), f is increasin on (, 1) ( 1,0), f is decreasin on (0,1) (1, ), x = 0 is a critical number of f, f (0) = 0 is a local maximum value of f. (b) Find the intervals on which the raph of y = f is concave upward or downward. Pae 4

5 Calculus Midterm Exam October 31, 018 f (x) = (x 1) ( 4) + 4x (x 1)x (x 1) 4 = 1x + 4 (x 1) 3 > 0 if x (, 1) (1, ), < 0 if x ( 1,1), the raph of f is concave upward on (, 1) (1, ), the raph of f is concave downward on ( 1,1). (c) Find the asymptotes, if it exists, of the raph y = f. and since Since x ± x x = = y = is the horizontal asymptoe, 1 x x 1 > x = for x > 1 = the raph of f lies above y = for x > 1 x x x x 1 ± x = and 1 x 1 ± x = ± = x = ±1 are vertical asymptotes of the raph of f. 1 (d) Sketch the raph of f. Solution: (, 1) ( 1, 0) (0, 1) (1, ) f +ve +ve ve ve f +ve ve ve +ve f concave upward concave downward concave downward concave upward Pae 5