5.1#3
The only critical point is at the middle, and is a (local) maximum.
5.1#4
The only critical point in in the middle, and is neither a local minimum nor maximum.
5.1#5
5.1#8
a. The demand for this product is increasing when demand' is positive, i.e. weeks 0-2 and 6-10.
b. Demand is at a local maximum after it's finished increasing and before it's started decreasing, so somewhere between weeks 2 and 3; call it 2 2/3 (but that can only be an estimate since we don't have enough information to know the exact time). Likewise, demand is at a local minimum after it's finished decreasing and before it's started increasing, so somewhere between weeks 5 and 6; call it 5 1/4 (again, only an estimate). These estimates are based on the assumption that demand' changes linearly from one week to the next, i.e. demand'' is constant. In sum: "2+something and 5+something".
5.1#11
5.1#13
The critical point happens when f'(x)=0. We know f'(x) = 2x + a. Since we want the critical point at x=-2, that means 0 = 2 (-2) + a, so a = 4.
At that value of x, though, we also want f(x)=-3. So -3 = x^2+ax+b = (-2)^2 + 4(-2) + b = -4 + b, so b = 1.
In sum: a=4, b=1.
5.1#15
Since f(x) = a x e^{bx}, f(1/3) = a/3 e^{b/3}. We want f(1/3)=1. In particular a=0 is impossible.
Meanwhile, we also want f'(1/3)=0. We know (again, product and chain rules) that f'(x) = a b x e^{bx} + a e^{bx} = a e^{bx} (bx + 1). Since a is not zero, nor e^{bx}, we must have bx+1=0 at x=1/3. So b=-3.
Plugging back in 1 = a/3 e^{b/3}, we get a = 3e.
In sum: a=3e, b=-3.
5.1#19
In all four cases: since there is only one critical point (at x=3), the derivative must have a constant sign before x=3, and also after x=3 (constant in each region - not necessarily constant the whole time). So if we find out that it's decreasing before or increasing after, it can't be a maximum; if we find out it's increasing before or decreasing after, it can't be a minimum.
a. f'(1)=3 increasing before, f'(5)=-1 decreasing after. It's a maximum.
b. f goes to infinity as x goes to +/- infinity. So it must decrease
before, and increase after. It's a minimum.
c. f(1)=1, f(2)=2 - increasing before. f(4)=4, f(5)=5 - increasing after.
It's neither a maximum nor a minimum.
d. f'(2)=-1 - decreasing before. f(3)=1. As x->infinity, f(x)->3 -
increasing after. It's a minimum.
20. f(x) = x^5 + x + 7 has exactly one real root. (By contrast, x^2-1 has two, and x^2+1 has none, just to give you an idea of the difficulty of the problem.)
First, why does it have one at all? As x -> -infinity, f(x) does too; likewise as x -> infinity, f(x) does too. But we can be more down-to-earth. f(0) = 7 which is positive, but f(-2) = -27 which is negative. So somewhere in between those two this continuous function must go to zero.
So why not more than once? Because f'(x) = 5x^4 + 1 > 0 for all x. This function is always increasing, so once it hits zero, it will only go higher (to the right) and lower (to the left) and never revisit zero. In the words of the hint: it has no critical points.
The Extra
The answers: f(x) = x^n has a local minimum at x=0 exactly if n is even. The only n for which f''(0)>0 is n=2.
To show the first part, we have to see that f(x) is minimized at x=0 for n even, but can go lower (for x arbitrarily close to 0) for n odd.
If n is even: n=2m. Then f(x)=(x^m)^2. Since it's a square, it's nonnegative, and so can't go less than f(0).
If n is odd: for any negative number x, f(x)<0. So that gets us arbitrarily close to x=0 while still being beaten by f(0).
(An argument that would NOT work: just pointing out that f(-1)=-1. That shows that f does not have a _global_ minimum at 0. For example, f(x) = x^2-x^4 does not have a global minimum at zero because f(2)