A curve has equation y=4x^3 -3x+3. Find the coordinates of the two stationary points. Determine whether each of the stationary points is a maximum or a minimum.

There are two stationary points

Note that 1/2 = 0.5

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How to get those answers:

Stationary points occur when the derivative is zero, when the graph is neither increasing nor decreasing (i.e. it's staying still at that snapshot in time).

Apply the derivative to get:

f(x) = 4x^3 - 3x + 3

f ' (x) = 12x^2 - 3

Then set it equal to zero and solve for x.

f ' (x) = 0

12x^2 - 3 = 0

12x^2 = 3

x^2 = 3/12

x^2 = 1/4

x = sqrt(1/4) or x = -sqrt(1/4)

x = 1/2 or x = -1/2

x = 0.5 or x = -0.5

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Let's do the first derivative test to help determine if we have local mins, local maxes, or neither.

Set up a sign chart as shown below. Note the three distinct regions A,B,C

The values -0.5 and 0.5 are not in any of the three regions. They are the boundaries.

The reason why I split things into regions like this is to test each region individually. We'll plug in a representative x value into the f ' (x) function.

To start off, we'll check region A. Let's try something like x = -2

f ' (x) = 12x^2 - 3

f ' (-2) = 12(-2)^2 - 3

f ' (-2) = 45

The actual result doesn't matter. All we care about is whether if its positive or negative. In this case, f ' (x) > 0 when we're in region A. This tells us f(x) is increasing on the interval

Let's check region B. I'll try x = 0.

f ' (x) = 12x^2 - 3

f ' (0) = 12(0) - 3

f ' (0) = -3

The result is negative, so f ' (x) < 0 when . The f(x) curve is decreasing on this interval.

The change from increasing to decreasing as we pass through x = -0.5 indicates that we have a local max here.

Plug x = -0.5 into the original function to find its paired y coordinate.

f(x) = 4x^3 - 3x + 3

f(-0.5) = 4(-0.5)^3 - 3(-0.5) + 3

f(-0.5) = 4

The point (-0.5, 4) is a stationary point. More specifically, it's a local max.

Side note: This is the same as the point (-1/2, 4) when written in fraction form.

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Let's check region C

I'll try x = 2

f ' (x) = 12x^2 - 3

f ' (2) = 12(2)^2 - 3

f ' (2) = 45

The positive outcome tells us that any number from region C does the same, and f ' (x) > 0 when . The function f(x) is increasing on this interval.

Region B decreases while C increases. The change from decreasing to increasing indicates we have a local min when x = 0.5

Plug this x value into the original equation

f(x) = 4x^3 - 3x + 3

f(0.5) = 4(0.5)^3 - 3(0.5) + 3

f(0.5) = 2

The local min is located at (0.5, 2) which is the other stationary point.

The graph and sign chart are shown below.

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step-by-step explanation: