HW #7 solutions

• sec. 4.2, p. 238, #4. f(x)=x*sqrt(x+6) is continuous on [-6,0] and differentiable on (-6,0) (though not at -6) since f'(x)=sqrt(x+6)+x/(2sqrt(x+6))=(3x+12)/(2sqrt(x+6)). But f(-6)=0=f(0), hence by Rolle's theorem there is a number c in (-6,0) with f'(c)=0, in fact c=-4 is the only c that works.
• #16 For f(x)=(x+1)/(x-1), f'(x)=-2/(x-1)^2 is always negative but (f(2)-f(0))/(2-0)=2 cannot be f'(c). This does not contradict the mean value theorem since f is not continuous on [0,2], it isn't even defined at 1.
• sec. 4.3, p. 247, #8 a) Given the graph of f'(x), we see that f'(x)>=0 and f is increasing on [2,4] and on [6,9]. b) f'(x) is zero at 0, 2, 4, and 6. At 2 and 6 f'(x) goes from negative to positive so these are local minima. At 4 f'(x) goes from positive to negative so this is a local maxima. At 0 f'(x) goes negative so we have an endpoint local maxima, while at 9 f'(x) is positive so we have another endpoint local maxima. c) The critical points of f'(x) are at 1, 3, 5, and 8. On [0,1], [3,5], and [7,8], f'(x) is decreasing so f is concave down. On [1,3], [7,7] and [8,9], f'(x) is increasing so f is concave up. d) The inflection points are where the concavity of f changes, in this case at all of the critical points of f', 1, 3, 5, and 8.
• #12 For f(x)=5-3x^2+x^3, we have f'(x)=-6x+3x^2 which is zero at 0 and 2, negative in between and positive outside this range. Hence a) f increases on (-infty,0] and [2,infty), and decreases on [0,2], b) f has a local max at 0 of 5 and a local min at 2 of 1. Since f''(x)=-6+6x is zero at 1, negative for x less than 1, and positive for x greater than 1, c) f is concave down on (-infty,1] and concave up on [1,infty) with an inflection point at 1.
• #14 For f(x)=x/(1+x)^2, we have f'(x)=(1-x)/(x+1)^3 which is zero at 1, undefined at -1, positive on (-1,1) and negative otherwise. Hence a) f is decreasing on (-infty,-1) and [1,infty), increasing on (-1,1] and has an absolute maximum value at 1 of 1/4, but no local minima. Since f''(x)=(2x-4)/(x+1)^4 is zero at 2, undefined at -1, negative for x less than 2 except -1, and positive for x greater than 2, c) f is concave up on [2,infty) and concave down on (-infty,-1) and (-1,2] with an inflection point at 2.
• #20 For f(x)=x^4(x-1)^3, f'(x)=x^3(x-1)^2(7x-4), so a) the critical points are 0, 1 and 4/7. b) Since f''(x)=6x^2(x-1)(7x^2-8x+2) has f''(0)=f''(1)=0 the second derivative doesn't tell you whether 0 or 1 are relative max's or min's. f''(4/7)=576/2401 is positive so f has a relative min at 4/7. c) Checking how the sign of f'(x) changes at each critical point we see f'(x) changes from positive to negative at 0 so 0 is a relative max, f'(x) changes from negative to positive at 4/7 so 4/7 is a relative min, and f'(x) is positive on both sides of 1 so 1 is a horizontal inflection point.
• #30 For g(x)=200+8x^3+x^4, we have g'(x)=24x^2+4x^3 is zero at 0 and -6, negative for x less than -6 and positive otherwise, and g''(x)=48x+12x^2 is zero at 0 and -4, negative between 0 and 4 and positive otherwise. Hence a) g(x) is increasing on [-6,infty), decreasing on (-infty,-6], b) has a local min at -6 of -232 but no max, c) is concave up on (-infty,-4] and [0,infty) but concave down on [-4,0] and has inflection points at -4, and at 0 (where the slope is zero). d) Graphing this we have
  
  
• #36 For Q(x)=x-3x^(1/3), Q'(x)=1-x^(-2/3) is zero at 1 and -1 and undefined at 0, negative between -1 and 1 except for 0, and positive otherwise. Q''(x)=2/3 x^(-5/3) is undefined at 0, positive for positive x and negative otherwise. Hence a) Q(x) is increasing on [-1,1], decreasing on (-infty,-1] and [1,infty), b) has a relative max at -1 of 2 and a relative min at 1 of -2, and c) is concave down on (-infty,0] and concave up on [0,infty). d) Graphing this we have
  
  
• sec. 4.4, p. 260, #10 lim x-> infty sqrt((2x^2-1)/(x+8x^2))= sqrt(lim (2-1/x^2)/(1/x+8))=sqrt((2-lim(1/x^2))/(8+lim(1/x)))=sqrt(2/8)=1/2
• #32 For y=(x^2+4)/(x^2-1) we have the denominator going to 0 at 1 and -1 so there are vertical asymptotes at x=-1 and x=1, and lim x-> infty (x^2+4)/(x^2-1)=lim(1+4/x^2)/(1-1/x^2)=1 so we have a horizontal asymptote at y=1. Graphing this we have