Math 3A Homework #8 solutions
5.3:12 Find the local and global extrema, and where the function is increasing and decreasing for y=e−x2/4,
x∈R.
Answer: With y′=−1
2xe−x2/4, we find one critical point at x=0. When x<0, y′>0, and when x>0,
y′<0. So the function is increasing left of zero, and deacreasing right of zero. Thus x=0 is a local
maximum. limx→∞f(x) = limx→−∞f(x) = 0, so x=0 is also a global maximum. Since the minima are
attained only at ±∞, there is no global minimum.
5.3:16 Find the local and global extrema, and where the function is increasing and decreasing for y=
√1+x2.
Answer: With y′=x
√1+x2, we find one critical point when x=0. Since the denominator is everywhere
positive, we find that the function is decreasing left of zero and increasing right of zero. Thus x=0 is a
local minimum. With limx→∞f(x) = limx→−∞f(x) = ∞, we see that the function has no global maxima, but
does have a global minimum at x=0.
5.3:20 Find the inflection points of f(x) = cosx, 0 ≤x≤π.
Answer: f′(x) = −sinxand f′′(x) = −cosx, so the concavity goes to zero when −cosx=0, which means
x=π
2. Since −cos(x)<0 when xis slightly less than π
2and −cos(x)>0 when xis slightly greater than π
2.
This means the concavity switches signs, and thus π
2is an inflection point.
5.3:24 Find the inflection points of f(x) = lnx+1
x,x>0.
Answer: f′(x) = 1
x−1
x2and f′′(x) = 2
x3−1
x2, so the concavity goes to zero when 2
x3=1
x2, which means
x=2. We see that when xis slightly less than 2, the function is concave up. And when x>2, the second
derivative is negative. So the concavity changes at x=2 and so x=2 is an inflection point.
5.3:36 Let
f(x) = −2
x2−1,x6=−1,1
Answer: Since xonly appears as an even power, the limits to ±∞are the same.
lim
x→±∞−2
x2−1=−lim
x→±∞
2/x2
1−1/x2=0
And now:
lim
x→−1−−2
x2−1=lim
x→−1−−2
(x−1)(x+1)
=−2
−1−1lim
x→−1−
1
1+x
=−∞
