MCV 4U1 Worksheet 2.2
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In Exercises 1-4, compare the right-hand and left-hand derivatives to In Exercises 5-10, the graph of a function over a closed interval D show that the function is not differentiable at the point P. is given. At what points does the function appear to be (a) differentiable? (b) continuous but not differentiable? (c) neither continuous nor differentiable? 5. 6.
7. 8.
9. 10.
In Exercises 11-16, the function fails to be differentiable at x = 0. In Exercies17-22, find all values of x for which the function is Tell whether the problem is a corner, a cusp, a vertical tangent, or a differentiable. discontinuity. MCV 4U1 Worksheet 2.2 Solutions
1. 2. 3. Left-hand derivative: Left-hand derivative: Left-hand derivative: f (0 h) f (0) f (1 h) f (1) f (1 h) f (1) lim lim lim h0 h h0 h h0 h 2 2 2 1 h 1 h 0 lim lim lim h0 h h0 h h0 h lim h lim 0 1 h 1 1 h 1 h0 h0 lim 0 0 h0 h 1 h 1 Right-hand derivative: Right-hand derivative: (1 h) 1 f (0 h) f (0) f (1 h) f (1) lim lim lim h0 h0 h h0 h h 1 h 1 h 0 2(1 h) 2 1 lim lim lim h0 h h0 h h0 1 h 1 lim 1 lim 2 1 h0 h0 2 1 2 Right-hand derivative: Since 0≠2, the function is not differentiable Since 0≠1, the function is not differentiable f (1 h) f (1) at the point P. at the point P. lim h0 h 2(1h) 1 1 lim h0 h 2h lim h0 h lim 2 h0 2 1 Since 2 , the function is not 2 differentiable at the point P.
4. 5(a).All points in [3, 2] 6(a). All points in [-2, 3] Left-hand derivative: f (1 h) f (1) (b). None (b). None lim h0 h (c). None (c). None (1 h) 1 lim h0 h lim 1 h0 1 Right-hand derivative: MCV 4U1 Worksheet 2.2 Solutions
f (1 h) f (1) lim h0 h 1 1 lim 1 h h0 h 1 (1 h) lim h0 h(1h) h lim h0 h(1h) 1 lim h0 1h 1 Since 1≠-1, the function is not differentiable at the point P.
7(a). All points in [-3, 3] except x = 0 8(a). All points in [-2, 3] except x = -1, 0, 2 9(a). All points in [-1, 2] except x = 0
(b). None (b). x = -1 (b). x = 0
(c). x = 0 (c). x = 0, x = 2 (c). None
10(a). All points in [-3, 3] except x = -2, 2 11. 12. Since lim tan1 x tan1 0 0 y(0) , y(0 h) y(0) (b). x = -2, x = 2 x0 lim h0 the problem is a discontinuity. h (c). None 4 h5 lim h0 h 1 lim h0 1 h5 y(0 h) y(0) lim h0 h 4 h5 lim h0 h 1 lim h0 1 h5 The problem is a cusp. MCV 4U1 Worksheet 2.2 Solutions
13. Note that 14. 15. 2 y(0 h) y(0) Note that y x x 2 lim h0 h y 3x 2 x 1 x x 2 3 3 h 3 5x 1, x 0 2, x 0 lim x 1, x0 2x 2, x0 h0 h y(0 h) y(0) y(0 h) y(0) 3 h lim lim lim h0 h h h0 h h0 (5h 1) ( 1) 2 2 lim 1 h0 h lim h0 h lim 2 h0 lim 5 lim 0 h 3 h0 h0 5 0 y(0 h) y(0) y(0 h) y(0) The problem is a vertical tangent. lim lim h0 h h0 h (h 1) ( 1) (2h 2) 2 lim lim h0 h h0 h lim 1 lim 2 h0 h0 1 2 The problem is a corner. The problem is a corner.
16. 17. 18. Find the zeros of the denominator. The function is differentiable except possibly 2 where 3x – 6 = 0, that is, at x = 2. We check x 4x 5 0 for differentiability at x = 2, using k instead (x 1)(x 5) 0 of the usual h, in order to avoid confusion x = -1 or x = 5 with the function h (x). The function is a rational function, so it is h(2 k) h(2) lim differentiable for all x in its domain: all reals k0 k except x = -1, 5. 3 3(2 k) 6 5 5 lim k0 k 3 3k lim k0 k 1 33 lim k0 2 k 3 The function has a vertical tangent at x = 2. It is differentiable for all reals except x = 2 MCV 4U1 Worksheet 2.2 Solutions
y(0 h) y(0) lim h0 h 3 h 0 lim h0 h 3 h lim h0 h 1 lim h0 2 h 3 y(0 h) y(0) lim h0 h 3 h 0 lim h0 h 3 h lim h0 h 1 lim h0 2 h 3 The problem is a cusp.
19. 20. 21. Note that the sine function is off, so Since the cosine function is even, so The function is piecewise-defined in terms of P() x Q() x polynomials, so it is differentiable everywhere except possibly at x = 0 and at x sin x 1 3cos x = 3. Check x = 0: sin x 1, x0 3cos x sin x 1, x 0 The function is differentiable for all reals. The graph of P(x) has a corner at x = 0. The function is differentiable for all reals except x = 0. MCV 4U1 Worksheet 2.2 Solutions
g(0 h) g(0) lim h0 h 2 (h 1) (1) lim h0 h 2 h 2h lim h0 h lim (h 2) h0 2 g(0 h) g(0) lim h0 h (2h 1) 1 lim h0 h lim 2 h0 2 The function is differentiable at x = 0. Check x = 3: Since g(3) = (4 -3)2 = 1 and lim g() x h3 lim (2x 1) h3 2(3) 1 7 The function is not continuous (and hence not differentiable) at x = 3. The function is differentiable for all reals except x = 3.
22. Note that C() x x x 2 x , x0 2 x , x 0 so it is differentiable for all x except possibly at x = 0. Check x = 0: MCV 4U1 Worksheet 2.2 Solutions
C(0 h) C(0) lim h0 h h h 0 lim h0 h lim h h0 0 The function is differentiable for all reals.