Respuesta :
Answer:
There are two stationary points
- Local max = (-0.5, 4)
- Local min = (0.5, 2)
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
- 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
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.
