1. f /(x) = 6x2 – 6x – 12 = 6(x2 – x – 2) = 6(x-2)(x+1).
Example 1 (a) Determine whether the function f(x) = 2x3 – 3x2 – 12x + 10 with domain [-2,3] has a minimum or a maximum value.
Solution Since f is a continuous function with domain a closed interval, f has minimum and maximum values by the Maximum Value Theorem. Observe that
f /(x) = 6x2 – 6x – 12 = 6(x2 – x – 2) = 6(x-2)(x+1).
Thus f has two critical points where its derivative is zero: x=2 and x=-1. In addition, the domain of f contains both of its endpoints: x=-2 and x=3. To find the minimum and maximum value of f we compare the values of f at these four points:
f(2) = -10, f(-1) = 17, f(-2) = 6, f(3) = 1.
Thus the minimum value of f is –10 which occurs at x=2 while the maximum value of f is 17 which occurs at x=-1. Note that both x=2 and x=-1 are interior points of the domain of f.
y = 2x3 – 3x2 – 12x + 10

2. (b) Determine whether the function g(x) = 2x3 – 3x2 – 12x + 10 with domain [-3,4] has a minimum or a maximum value.
Solution Since g is a continuous function with domain a closed interval, g has minimum and maximum values by the Maximum Value Theorem. Note that g has the same critical points as the function f of (a): x=2 and x=-1. To find the minimum and maximum values of g we compare the values of g at these points and the endpoints x=-3 and x=4 of the domain of g:
g(2) = -10, g(-1) = 17, g(-3) = -35, g(4) = 42.
Thus the minimum value of g is –35 which occurs at x=-3 while the maximum value of g is 42 which occurs at x=4. Note that the minimum and maximum values of g occur at the endpoints of its domain.
y = 2x3 – 3x2 – 12x + 10

3. h(2) = -10, h(-1) = 17, h(-3) = -35, h(3) = 1.
(c) Determine whether the function h(x) = 2x3 – 3x2 – 12x + 10 with domain [-3,3] has a minimum or a maximum value.
Solution Since h is a continuous function with domain a closed interval, h has minimum and maximum values by the Maximum Value Theorem. Note that h has the same critical points as the function f of (a): x=2 and x=-1. To find the minimum and maximum value of h we compare the values of h at these points and at the endpoints x=-3 and x=3 of its domain:
h(2) = -10, h(-1) = 17, h(-3) = -35, h(3) = 1.
Hence h has its minimum value of –35 at its left endpoint x=-3 and its maximum value of 17 at the critical point x=-1.
y = 2x3 – 3x2 – 12x + 10

4. (d) Determine whether the function k(x) = 2x3 – 3x2 – 12x + 10 with domain (-3,4) has a minimum or a maximum value.
Solution Since the domain of k is not a closed interval, the Maximum Value Theorem does not apply to k, and we do not know before we investigate whether k has a minimum or maximum value. Note that k has the same critical points as the function f of (a): x=2 and x=-1. Since the domain of k does not contain its endpoints, we compute the limit of k(x) as x approaches each of these endpoints.
Since the first limit is less than k(2)=-10 and the second limit is greater than k(-1)=17, it follows that k has neither a minimum nor a maximum value.
y = 2x3 – 3x2 – 12x + 10

5. (e) Determine whether the function m(x) = 2x3 – 3x2 – 12x + 10 with domain (-3,4] has a minimum or a maximum value.
Solution Since the domain of m is not a closed interval, the Maximum Value Theorem does not apply to m, and we do not know before we investigate whether m has a minimum or maximum value. Note that m has the same critical points as the function f of (a): x=2 and x=-1. The domain of m contains the right
endpoint x=4 of (-3,4] . We compare the values of m at these three points with the limit of m(x) as x approaches the left endpoint x=-3 of (-3,4] which is not in the domain of h.
m(2) = -10, m(-1) = 17, m(4) = 42,
We see that m has maximum value 42 at the right endpoint of its domain x=4. Since the value -35 of the above limit is less that the value –10 of m at the local minimum at x=2, it follows that m does not have a minimum value.
y = 2x3 – 3x2 – 12x + 10