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Chapter 15: Problem 339

Determine the critical points of \(\mathrm{f}(\mathrm{x})=3 \mathrm{x}^{4}-4 \mathrm{x}^{3}\) and sketch the graph.

Short Answer

Expert verified
The critical points of \(f(x) = 3x^4 - 4x^3\) are at \(x = 0\) and \(x = 1\). The second derivative test indicates a local minimum at \(x = 1\), while the test is inconclusive for \(x = 0\). The graph of the function displays an overall rise, reaching a critical point at \(x = 0\), decreasing to its local minimum at \(x = 1\), and then rising again.

Step by step solution

01

Finding the first derivative

To find the critical points of the function, we first need to find its derivative. Differentiate the function \(f(x) = 3x^4 - 4x^3 \) with respect to \(x\): \( f'(x) = \frac{d}{dx}(3x^4 - 4x^3) = 12x^3 - 12x^2 \)
02

Setting the first derivative equal to zero

Now, to find the critical points, we set the first derivative equal to zero and solve for \(x\): \( 12x^3 - 12x^2 = 0 \) Factor out the common term, \(12x^2\): \( 12x^2(x - 1) = 0 \) The function has critical points when \(x = 0\) and \(x = 1\).
03

The second derivative test

To determine the nature of the critical points (local minima, local maxima, or saddle points), we'll use the second derivative test. First, we find the second derivative of the function: \( f''(x) = \frac{d^2}{dx^2}(12x^3 - 12x^2) = 36x^2 - 24x \) Now, we evaluate the second derivative at the critical points: 1. For \(x = 0\): \(f''(0) = 36(0)^2 - 24(0) = 0\). The second derivative test is inconclusive for this point. 2. For \(x = 1\): \(f''(1) = 36(1)^2 - 24(1) = 12 > 0\). Since the second derivative is positive, this indicates a local minimum at \(x = 1\).
04

Sketch the graph

To sketch the graph, we now have the following information: 1. The function has a local minimum at \(x = 1\). 2. The function also has a critical point at \(x = 0\), but the second derivative test is inconclusive. Considering this information, we can roughly sketch the graph of the function \(f(x) = 3x^4 - 4x^3\). The function rises from negative infinity and reaches a critical point at \(x = 0\). Then, it decreases until it reaches a local minimum at \(x = 1\), after which it rises again to positive infinity. To complete the sketch, we can plot the function using graphing software or a calculator, incorporating the critical points and local minimum found in the previous steps.

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