x 2 x 2 cos 1 x x2, lim 1. If x > 0, multiply all three parts by x > 0, we get: x x cos 1 x x, lim lim x cos 1 lim = 5 lim sin 5x

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1 Homework 4 3.4,. Show that x x cos x x holds for x 0. Solution: Since cos x, multiply all three parts by x > 0, we get: x x cos x x, and since x 0 x x 0 ( x ) = 0, then by Sandwich theorem, we get: x 0 x cos x = 0. b. Use the sandwich theorem to show that x 0 x cos x = 0. Solution: Since cos x. If x > 0, multiply all three parts by x > 0, we get: x x cos x x, and since x 0+ x x 0+ ( x) = 0, then by Sandwich theorem, we get: If x < 0,multiply all three parts by x < 0, we get: x cos x 0+ x = 0. x x cos x x, and since x 0 x x 0 ( x) = 0, then by Sandwich theorem, we get: x 0 x cos x = 0. Therefore, we get x cos x 0 x = 0. sin 5x sin 5x 5 x 0 x x 0 5x = 5 sin 5x x 0 5x = 5. sin x x 0 x = sin x x 0 x =. 3.5, 5. Use the intermediate-value theorem to show that e x = x has a solution in (0, ). Solution: Let f(x) = e x x, then f(0) = > 0, and f() = e < 0, thus by I-V theorem, there is c in (0, ), such that f(c) = 0, that is e c = c. 6. Use the intermediate-value theorem to show that cos x = x has a solution in (0, ). Solution: Let f(x) = cos x x, then f(0) = > 0, and f() = cos < 0, thus by I-V theorem, there is c in (0, ), such that f(c) = 0, that is cos c = c. 3 Explain why a polynomial of degree 3 has at least one root. Solution: Suppose f(x) = ax 3 + bx + cx + d is a polynomial of degree 3, where a, b, c, d are constants. If a > 0 then, there is a number t 0 big enough, such that f(t) > 0, and there is a number s 0 small

2 enough, such that f(s) < 0, therefore by I-V theorem, we get there is l in (s, t), satisfies that f(l) = 0, that is f(x) has at least one root l. Similarly, if a < 0 then, there is a number t 0 big enough, such that f(t) < 0, and there is a number s 0 small enough, such that f(s) > 0, therefore by I-V theorem, we get there is l in (s, t), satisfies that f(l) = 0, that is f(x) has at least one root l. 4 Explain why a polynomial of degree n, where n is an odd number has at least one root. Solution: Suppose f(x) = a n x n + a n x n + + a x + a x + a 0 is a polynomial of degree n, where a n, a n,, a 0 are constants. Because n is odd, if a n > 0 then, there is a number t 0 big enough, such that f(t) > 0, and there is a number s 0 small enough, such that f(s) < 0, therefore by I-V theorem, we get there is l in (s, t), satisfies that f(l) = 0, that is f(x) has at least one root l. Similarly, if a n < 0 then, there is a number t 0 big enough, such that f(t) < 0, and there is a number s 0 small enough, such that f(s) > 0, therefore by I-V theorem, we get there is l in (s, t), satisfies that f(l) = 0, that is f(x) has at least one root l. 5. Explain why y = x 4 has at least two roots. Solution: Since y(3) = 5 > 0, and y() = 3 < 0, then by I-V theorem, there is a number c in (, 3), such that y(c ) = c 4 = 0. Also since y( 3) = 5 > 0, and y( ) = 3 < 0, then by I-V theorem, there is a number c in ( 3, ), such that y(c ) = c 4 = 0, and because c and c are in disjoint sets respectively, so they are different, therefore y = x 4 has at least roots c, c. 4., 5. Use the formal definition to find the derivative of y = x, for x > 0. Solution: Denote f(x) = x, then by the formal definition of derivative, for x > 0, we have f f(x + h) f(x) x + h x ( x + h x)( x + h + x) (x) h h h( x + h + x) x + h x h( x + h + x) = x + h + x x. 6. Use the formal definition to find the derivative of f(x) = x+ for x. Solution: Denote f(x) = x, then by the formal definition of derivative, for x, we have f f(x + h) f(x) (x) h h h(x + + h)(x + ) x++h x+ x+ (x++h) (x++h)(x+) h h (x + + h)(x + ) = (x + ). 7. Find the equation of the tangent line to the curve y = 3x at the point (, 3). Solution: y (x) = 6x, then y () = 6, which is the slope of the tangent line. Since the tangent line pass the point (, 3), thus the equation of the tangent line is: y 3 = 6(x ), that is y = 6x Find the equation of the tangent line to the curve y = /x at the point (, ). Solution: y (x) = x, then y () =, which is the slope of the tangent line. Since the tangent line pass the point (, ), thus the equation of the tangent line is: y = (x ), that is y = x Which of the following statements is true: (A) If f(x) is continuous, then f(x) is differentiable. (B) If f(x) is differentiable, then f(x) is continuous. Solution: (A) is false. Counterexample: f(x) = x /3 is continuous at x = 0, but not differentiable at x = 0. (B) is true, the proof is in page 4-4.

3 5.Explain the relation between continuity and differentability. Solution: If f is differentiable at x = c, then f is also continuous at x = c; but if f is continuous at x = c, then f need not be differentiable at x = c. 4., 7. Differentiate the function g(s) = 5s 7 + s 3 5s. Solution: g (s) = 5 7s 6 + 3s 5 = 35s 6 + 6s 5. 8.Differentiate the function g(s) = 3 4s 4s 3. Solution: g (s) = 4 s 4 3s = 8s s. 9. Differentiate f(x) = x 3 rx s with respect to x. Assume that r and s are constant. Solution:f (s) = 3x r = 3 x r. 30 Differentiate f(x) = r+x x + (r + s)x with respect to x. Assume that r and s are nonzero constants. Solution: f (x) = ( r + x x + (r + s)x) = + r + s. 8. Suppose that P (x) is a polynomial of degree 4. Is P (x) a polynomial as well? If yes, what is the degree. Solution: Yes. Because if P (x) = ax 4 + bx 3 + cx + dx + e, where a, b, c, d, e are all constants. then P (x) = 4ax 3 + 3bx + cx + d, therefore degree of P (x) is Suppose that P (x) is a polynomial of degree k. Is P (x) a polynomial as well? If yes, what is the degree? Solution: Suppose P (x) = a n x n + a n x n + + a x + a x + a 0, then P (x) = na n x n + (n )a n x n + + a x + a, thus the degree of P (x) is n. 4.3, 3. Differentiate f(x) = a(x a) + a with respect to x. Assume that a is a positive constant. Solution: Use the product rule to f(x) = a(x a)(x a) + a: f (x) = a x (x ) + a x (x ) = 8ax(x a) = 8ax 3 8a x. 3 Differentiate f(x) = 3(x ) +a with respect to x. Assume that a is a positive constant. Solution: Use the product rule to f(x) = 3 +a (x )(x ), we get: Differentiate the following functions: 67.g(s) = s/3 s /3. Solution: g (s) = 68.gs = s/7 s /7 s 3/7 +s 4/7. Solution: f (x) = 3 6 ((x ) + (x )) = (x ) + a s /3 (s /3 ) 3 (s/3 )s /3 (s /3 ) = 3 3 s / s /3 (s /3 ) = 3 3 s /3 + 3 s /3 (s /3 ). g (s) = ( 7 s 6/7 7 s 5/7 )(s 3/7 + s 4/7 ) (s /7 s /7 )( 3 7 s 4/ s 3/7 ) (s 3/7 + s 4/7 ) = 7 s 3/7 s /7 + s /7 (s 3/7 + s 4/7 ) 3

4 Assume that f(x) is differentiable. Find an expression for the derivative of y at x =, assuming that f() = and f () = : 84. y = f(x) x +. Solution: y (x) = f (x)(x +) xf(x) (x +), then: y () = f ()( + ) 4f() ( + ) = = y = x +4f(x) f(x) Solution:y (x) = (x+4f (x))f(x) (x +4f(x))f (x) (f(x) ), thus: y () = (4 + 4f ())f() ( + 4f())f () (f() ) = 8 ( ) 0 = 8. 4

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