Find the absolute maximum and minimum values of the function

Question

a) f(x) = x³-3x²-9x+1 on the interval [−2,0]

b) g(x) = sin(2x) + 2 cos(x) on the interval [0,π/2].

Answer 1

We need to find the derivative of f(x), which is 3x² - 6x - 9, which is the same as 3(x² - 2x - 3). The minimum then is when x² - 2x - 3 = 0. Since we can rewrite it as (x-3)(x+1) = 0, the possible maxima/minima must occur at either x=3, x=-1, or the endpoints of the interval [-2, 0].

Since x=3 is outside the interval, we can ignore it.

For x=-1, look at the second derivative: 6x - 6. At the x=-1 value we found, we can see that the second derivative is negative, indicating that the function is concave down, meaning that it is a local maximum. Since f(0) < f(-1) and f(-2) < f(-1), then the value of x=-1 is the absolute maximum in the interval [-2, 0].

So what is the absolute minimum? Since there are no other places in the interval where the derivative is 0, then the minimum must be at one of the endpoints. f(0) = 1, and f(-2) = -1, so x=-2 is the absolute minimum.

For the other one g'(x) = 2(cos(2x) - sin(x)), and g''(x) = -2(sin(2x) + cos(x)). I encourage you to investigate this problem in a similar way.