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f(1) [3] 8. Let f(x) = 2x2 +3.r. Use the definition of the derivative to prove that f'(x) = 4x+3. Solution: We have f(x) - f(a) 2,c2 + 3.r - 2a? - 3a 1+ I -a 1 r - a - lim 2(x + a)(x - a) + 3(x -a) -a = lim 2(+ a) +3 = 4a +3 = lim a 1-a - as required.

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Lelia Lubowitz
Lelia LubowitzLv2
26 Sep 2018
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11. Let f(1) = cos 1. Use the definition of the derivative to prove that f'(:1) = -sin . Solution: We have · f'(:1) = lim cos(2+h) - cos(2) +0 cos(1) cos(h) - sin() sin(h) - cos(1) = lim h0 sin / – Jin contar) (cos(1) = -1) = sin(e) (sinm) h0 sin(h) cos(h) - = lim cos(2) h cos(h) - 1 = cos(x) lim h- oh = cos(x)0 - sin(x)(1) = - sin h c) lim 0 / h

[8] 9. Either prove the statement is true or show that it is false. (a) sin (sin(2)) = 27. Solution: It is FALSE. The range of sin-' r is (-1,1], hence sin (sin(27)) = sin (0) = 0 (b) If f is continuous at a, then f is continuous at a. Solution: It is TRUE. Let g(x) = \f()]. If lim f(x) = f(a), then we have since the absolute value function is continuous lim g(x) = lim f() = lim f(x) = f(a) = g(a) So, g = \f is continuous at a. (c) If [g] is continuous at a, then g is continuous at a. -1 if r <0 Solution: It is FALSE. Let g(x) = 3 · Then, g = 1, so g is continuous at 0 i if >0 by the Continuity Theorems, but lim g(t) = -1 + lim g(1) 240- 1- 0+ So, g(x) is not continuous at a. (d) If f is continuous at a, then f is differentiable at a. Solution: It is FALSE. We know that f(1) = 2 is continuous at 0, but is not differentiable at 0.

[3] 10. Prove the theorem that states: If f is differentiable at a, then f is continuous at a. Solution: Assume lim (*)={@) = f'(a). Then, since lim (1 – a) = 0, we get by the Limit Laws that f(x) - - @ -a) = lim + f(x)-f(a) T-a lim( 1a -a) = f'(a)-0 = 0 ma a limf (x) - f(a)] = lim- rta Now, observe that lim f(1) = lim[f (2) - f(a) + f(a)] = lim(f(x) – f(a)] + lim f(a) = 0 + f(a) = f(a) Hence, f(x) is continuous at a.

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