maximum and/or minimum valuse

Question

For each of the following functions defined on the indicated interval, find the absolute maximum and/or minimum values.

a) f(x)=2x³+3x²+4; -2 is less than or equal to x is less than or equal to ¹

b) f(x)=x^3-3x+1; 0 is less than or equal to x is less than or equal to 2

Answer 1

To find the max/min on an interval:

1. Take the derivative and set it equal to zero to find local extremes
2. Check the endpoints to see if they are higher/lower than the local extremes
3. Double check for any vertical asymptotes on the interval

For a), the derivative is 6x² + 6x and setting that equal to zero we get:

6x² + 6x = 0

6x(x + 1) = 0

x = 0 and x = -1 since both of those are in the interval, they are possibilities to be absolute min/max values

f(0) = 2(0)³ + 3(0)² + 4 = 4

f(-1) = 2(-1)³ + 3(-1)² + 4 = 5

Now we need to check the endpoints:

f(-2) = 2(-2)³ + 3(-2)² + 4 = 0

f(1) = 2(1)³ + 3(1)² + 4 = 9

There are no vertical asymptotes on the interval because the derivative is not undefined on any point in the interval.

So to summarize:

f(-2) = 0

f(-1) = 5

f(0) = 4

f(1) = 9

Meaning that, on the interval [-2, 1], the function values start at 0, go up to 5, go back down to 4, then go back up to 9 so the absolute min is 0 (occurring at x = -2) and the absolute max is 9 (occurring at x = 1)

Use the same process for the function noted in b).