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Find the \(x\) -coordinates of all critical points of the given function. Determine whether each critical point is a relative maximum, minimum, or neither by first applying the second derivative test, and, if the test fails, by some other method. \(f(x)=3 x^{4}-2 x^{3}\)

Short Answer

Expert verified
The critical points for the given function occur at \(x = 0\) and \(x = \frac{1}{2}\). After applying the second derivative test and first derivative test, we found that \(x = 0\) is a relative maximum, and \(x = \frac{1}{2}\) is a relative minimum.

Step by step solution

01

Find the first derivative

To find the first derivative of the function, apply the power rule to each term. \(f'(x) = \frac{d}{dx}(3x^4 - 2x^3) = 12x^3 - 6x^2\) Next, set the first derivative equal to zero and solve for x: \(0 = 12x^3 - 6x^2\)
02

Solve for x-coordinates of critical points

Factor the equation from the previous step to solve for x: \(0 = 6x^2(2x - 1)\) This implies that the critical points occur at \(x = 0\) and \(x = \frac{1}{2}\).
03

Apply the second derivative test

Find the second derivative of the function, which involves finding the derivative of the first derivative: \(f''(x) = \frac{d^2}{dx^2}(12x^3 - 6x^2) = 36x^2 - 12x\) Now, plug the x-coordinates of the critical points into the second derivative: \(f''(0) = 36(0)^2 - 12(0) = 0\) \(f''\left(\frac{1}{2}\right) = 36\left(\frac{1}{2}\right)^2 - 12\left(\frac{1}{2}\right) = 6\) Since the second derivative test is inconclusive for \(x = 0\) (as \(f''(0) = 0\)), we will need to apply another method. However, for \(x = \frac{1}{2}\), the second derivative test tells us that it's a relative minimum point, because \(f''\left(\frac{1}{2}\right) > 0\).
04

Apply the first derivative test for x = 0

Analyze the sign of the first derivative near \(x = 0\): For \(x < 0\), \(f'(x) = 12x^3 - 6x^2 > 0\), since \(x^3 > x^2\) for negative x values. For \(x > 0\), \(f'(x) = 12x^3 - 6x^2 < 0\), since \(x^3 < x^2\) for positive x values between 0 and 1. Therefore, the function transitions from increasing to decreasing at \(x = 0\), and this critical point is a relative maximum.
05

Write the conclusion

There are two critical points for the given function: - \(x = 0\) is a relative maximum, as determined by the first derivative test. - \(x = \frac{1}{2}\) is a relative minimum, as determined by the second derivative test.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Derivative Test
To identify critical points and analyze how a function behaves at those points, the first derivative test is an essential tool. It focuses on the change of direction in the slope of the function's graph. When the first derivative of the function, represented as \(f'(x)\), transitions from positive to negative, the function is said to have a relative maximum at that point. Conversely, if \(f'(x)\) changes from negative to positive, the point is a relative minimum.

In the exercise, after factoring the first derivative to find \(x = 0\) and \(x = \frac{1}{2}\), we observed the sign change of \(f'(x)\) around those points. For \(x = 0\), since \(f'(x)\) switched from positive to negative, it indicated a relative maximum. This change in the sign of the slope, detailed in step 4, is what confirms the behavior at that critical point without needing the second derivative test.

It's critical to examine the intervals around the critical points when applying this test. If the derivative does not change sign, then the critical point does not represent a relative extremum.
Second Derivative Test
While the first derivative test uses the sign of the derivative to infer the nature of critical points, the second derivative test uses the curvature of the function, as indicated by the sign of the function's second derivative, \(f''(x)\). This test can quickly determine whether a critical point is a concave up (relative minimum) or concave down (relative maximum).

For the function in our exercise, the second derivative was \(f''(x) = 36x^2 - 12x\), in which we substituted the critical points to evaluate it. At \(x = \frac{1}{2}\), the second derivative is positive, indicating a relative minimum because the function is concave up. However, at \(x = 0\), \(f''(0) = 0\), rendering the second derivative test inconclusive; a different approach—like the first derivative test—is required in such cases.

The second derivative test is favored for its simplicity, but it’s important to be aware of its limitations and be prepared to use the first derivative test when needed.
Relative Extrema
Relative extrema refer to the high and low points on a function within a specific interval. These points are categorized as relative maxima or minima based on their heights compared to nearby points on the function. Identifying relative extrema is crucial in understanding the overall shape and behavior of functions.

In the context of the exercise, the function displayed two critical points, one of which was a relative maximum at \(x = 0\), and the other a relative minimum at \(x = \frac{1}{2}\). These findings derive from comprehensively applying the first and second derivative tests. Recognizing these extrema proves valuable in scenarios like optimizing real-life systems or examining the turning points on a graph.

When determining relative extrema, it's imperative to examine the function comprehensively in the neighborhood of critical points to correctly infer the behavior and ensure the precise classification of each extremum.

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Most popular questions from this chapter

A production formula for a student's performance on a difficult English examination is given by $$g=4 h x-0.2 h^{2}-10 x^{2}$$ where \(g\) is the grade the student can expect to obtain, \(h\) is the number of hours of study for the examination, and \(x\) is the student's grade point average. The instructor finds that students' grade point averages have remained constant at \(3.0\) over the years, and that students currently spend an average of 15 hours studying for the examination. However, scores on the examination are dropping at a rate of 10 points per year. At what rate is the average study time decreasing?

A gencral cxponcntial demand function has the form \(q=A e^{-b p}(A\) and \(b\) nonzero constants). a. Obtain a formula for the price elasticity of demand at a unit price of \(p\) b. Obtain a formula for the price that maximizes revenue.

If we regard position, \(s\), as a function of time, \(t\), what is the significance of the third derivative, \(s^{\prime \prime \prime}(t) ?\) Describe an everyday scenario in which this arises.

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