MATH 180 FALL 1995

ASSIGNMENT #29

5.1, #5. f(x) = 2x^{3} + 3x^{2} − 36x + 5, f′(x) = 6x^{2} + 6x − 36 = 6(x + 3)(x − 2). The critical points −3 and x = 2 divide the domain of f into three intervals: (−∞, −3), (−3, 2), (2, ∞). The sign of f′ on each interval can be determined by evaluating f′(x) = 6(x + 3)(x − 2) at a suitable point, say −5 in (−∞, −3), 0 in (−3, 2), 3 in (2, ∞). We find f′(−5) = 6(−2)(−7) > 0, f′(0) = 6(3)(−2) < 0, f(3) = 6(6)(1) > 0. So x = −3 is a local maximum of f and x = 2 is a local minimum of f by the first-derivative test.

5.1, #9. f(x) = \frac{x}{x^{2} + 1}, f′(x) = \frac{(x^{2} + 1) · 1 − x · 2x}{(x^{2} + 1)^{2}} = \frac{1 − x^{2}}{(x^{2} + 1)^{2}} . f′(x) is defined

everywhere and is zero at x = −1 and 1. The sign of f′(x) is the sign of the numerator 1 − x^{2}, since the denominator (x^{2} + 1)^{2} is always positive. 1 − x^{2} is negative on (−∞, −1), positive on (−1, 1), and negative on (1, ∞). Thus x = −1 is a local minimum of f and x = 1 is a local maximum of f.

5.1, #13. f(x) = x − ln x, f′(x) = 1 − (1/x) = (x − 1)/x. Thus f has one critical point x = 1 on the given domain [0.1, 2]. The sign of f′(x) is negative on [0.1, 1), positive on (1, 2]. Thus x = 1 is a local minimum. For global extrema we compute f(0.1) = 2.40, f(1) = 1, f(2) = 1.31. Thus f has a global maximum at x = 0.1 and a global minimum at x = 1.

5.2, #1. The inflection points of f are the points where f′′ changes sign, or equivalently, where f′ has a local extremum. So the tops and bottoms of the graph of f′ are inflection points of f.

5.2, #4. y = 0.5xe^{−10x}, y′ = (\frac{1}{2} − 5x)e^{−10x}, y′′ = (50x − 10)e^{−10x}. The root x = 1/10

of y′ is the critical point of y. The sign of y′ is that of (1/2) − 5x since e^{−10x} is always positive. Thus x = 1/10 is a local maximum of y by the first-derivative test. The root x = 1/5 of y′′ is an inflection point of y since y′′ changes sign at x = 1/5.

5.2, #7. f(x) = sin x + cos x, f′(x) = cos x − sin x, f′′(x) = − sin x − cos x. The critical points of f in the interval [0, 6π] are x = π/4, 5π/4, 9π/4, 13π/4, 17π/4, 21π/4. The graphs of sin x and cos x show f′(x) changes sign at these critical points. The points alternate between local maxima and local minima with x = π/4 being a local maximum. The zeros of f′′(x) in [0, 6π] are x = 3π/4, 7π/4, 11π/4, 15π/4, 19π/4, 23π/4. These are inflection points since f′′(x) changes sign at these points.