# Find the absolute maximum and absolute minimun, if they exist, for `y=f(x)=2x^3 -3x^2 -12x+24`    A). `[-3,4]` B). `[-2,3]`

txmedteach | High School Teacher | (Level 3) Associate Educator

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To find the absolute minima and maxima for a function over a given interval, you need to check two categories of points:

1) At the boundaries of the interval

2) At points where `(df)/(dx) = 0`

For both, let's check the values at the boundaries. For A, this means checking `f(-3)` and `f(4)`, and for B, this means checking `f(-2)` and `f(3)`.

`f(-3) = 2(-3)^3-3(-3)^2-12(-3) + 24 = -21`

`f(4) = 56`

`f(-2) = 20`

`f(3) = 15`

Now that we have established the function values at the boundaries, we must now solve for where the function has derivatives equal to zero. Start by taking the derivative:

`(df)/(dx) = 6x^2-6x-12`

Now, solve for the x-values at which the derivative is zero:

`0 = 6x^2-6x-12`

`0 = x^2-x-2`

We can solve this equation for `x` by easily factoring:

`0 = (x-2)(x+1)`

This result gives us the two `x` values at which we might find other minima or maxima: `x = 2` and `x = -1`.

Now, we find the function values at `2` and `-1`:

`f(-1) = 31`

`f(2) = 4`

Both of these values are present on both intervals, so we need to take both into account when finding extrema.

Solving for the absolute minima and maxima for A, we simply find which `x` value gives the smallest and largest values for `f(x)` from the `x`-values `-3`, `-1`, `2`, and `4`. Clearly, the maxima and minima are found at the boundaries. `(-3, -21)` is the absolute minimum, and `(4, 56)` is the absolute maximum.

For B, we must do the same, but now we consider the function values at `x =` `-2`, `-1`, `2`, and `3`. Now, the absolute minimum and maximum are not at the boundaries. Instead, we see the absolute minimum at `(2, 4)` and the absolute maximum at `(-1, 31)`.

Recapping:

A) Minimum: (-3, -21); Maximum: (4, 56)

B) Minimum: (2, 4); Maximum: (-1, 31)

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