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.

Answer:

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 [tex]-\infty < x < -0.5[/tex]

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 [tex]-0.5 < x < 0.5[/tex]. 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 [tex]0.5 < x < \infty[/tex]. 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.

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

Stationary points occur when the derivative is zero, when the graph is neither increasing nor decreasing.

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 = 1/2 or x = -1/2

x = 0.5 or x = -0.5

The first derivative test to help determine if we have local mins, local maxes, or neither.

A = numbers to the left of -0.5

B = numbers between -0.5 and 0.5 excluding both endpoints

C = numbers to the right of 0.5

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

Let's try something like x = -2

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

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

f ' (-2) = 45

In this case, f ' (x) > 0 when we're in region A.

Let's check region B x = 0.

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

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

f ' (0) = -3

The result is negative, so f ' (x) < 0 when 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.

Substitute 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, it's a local max.

Let's check region C 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.

Susbtitute 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

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

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