Question

A function \(f(x)\) is given. Find the \(x\) values where \(f^{\prime}(x)\) has a relative maximum or minimum. \(f(x)=\frac{x^{4}}{4}+\frac{x^{3}}{3}-2 x+3\)

Short answer

The first derivative has a relative minimum at \(x = -\frac{2}{3}\).

Step-by-step solution

Find the first derivative

To find where the first derivative has a relative maximum or minimum, we first need to calculate the derivative of the function. Given the function \( f(x) = \frac{x^4}{4} + \frac{x^3}{3} - 2x + 3 \), we find the derivative:\[ f'(x) = x^3 + x^2 - 2 \].

Find the second derivative

Next, we find the second derivative of the function, which will help us determine the concavity of the first derivative. Taking the derivative of \( f'(x) = x^3 + x^2 - 2 \), we get:\[ f''(x) = 3x^2 + 2x \].

Find critical points of the first derivative

To find where \( f'(x) \) has relative extrema, we find its critical points by setting \( f''(x) = 3x^2 + 2x = 0 \) and solving for \( x \):\[ 3x(x + \frac{2}{3}) = 0 \].This gives the solutions \( x = 0 \) or \( x = -\frac{2}{3} \).

Determine nature of critical points

We use the second derivative to determine the nature of the critical points of \( f'(x) \). We find the sign of \( f''(x) \) at \( x = 0 \) and \( x = -\frac{2}{3} \):- For \( x = 0 \), \( f''(x) = 0 \).- For \( x = -\frac{2}{3} \), check a test point slightly greater (e.g., \( x = -\frac{1}{2} \) and slightly less \( x = -1 \)), to determine concavity. In this case, \( f''(x) \) changes sign from negative to positive, indicating \( x = -\frac{2}{3} \) is where \( f'(x) \) has a relative minimum. For \( x = 0 \), the test yields no sign change as both sides produce a positive value, showing no relative max/min.

Key concepts

First Derivative

In calculus, the first derivative of a function, typically denoted as \( f'(x) \), represents the rate of change of the function \( f(x) \) at any point \( x \). It's essentially the slope of the tangent line to the curve at that point. Finding the first derivative is the essential first step to understanding how a function behaves and evolves. In the case of polynomial functions, such as \( f(x) = \frac{x^4}{4} + \frac{x^3}{3} - 2x + 3 \), the first derivative \( f'(x) \) reveals where the function \( f(x) \) is increasing or decreasing. By applying rules of differentiation—such as the power rule—we derive the expression \( f'(x) = x^3 + x^2 - 2 \). This tells us how the original function changes with small changes in \( x \). It's a key tool for finding critical points, which can indicate maximum or minimum values of the function or places where the function doesn't have a steep slope.

Second Derivative

The second derivative of a function, \( f''(x) \), provides us with valuable insights about the behavior of \( f'(x) \). Essentially, \( f''(x) \) tells us how the rate of change (the first derivative) itself is changing. When the second derivative is positive, the first derivative is increasing, which means the original function is concave up (shaped like a cup); when \( f''(x) \) is negative, \( f'(x) \) is decreasing, and the function is concave down (shaped like a cap). In our specific problem, after finding the first derivative \( f'(x) = x^3 + x^2 - 2 \), we calculated the second derivative \( f''(x) = 3x^2 + 2x \) to help determine the concavity. This is crucial for identifying the relative extrema of \( f'(x) \), as it helps us discern if points like \( x = 0 \) or \( x = -\frac{2}{3} \) are indeed minima, maxima, or neither.

Critical Points

Critical points occur where the first derivative \( f'(x) \) is zero or undefined. These are potential locations for relative maxima or minima of the original function, as the slope change indicates possible peaks or troughs. In our exercise, solving the equation \( f''(x) = 3x^2 + 2x = 0 \) reveals the critical points, which are natural points of interest. By factoring \( 3x(x + \frac{2}{3}) = 0 \), we find that the critical points for the function are \( x = 0 \) and \( x = -\frac{2}{3} \). These critical points suggest places where the derivative \( f'(x) \) may have reached a maximum or minimum, though additional verification is needed to determine their nature.

Relative Extrema

Relative extrema are the points in the domain of a function where it reaches a local maximum or minimum. To classify these points, we use the second derivative test. This test involves checking the sign of \( f''(x) \) at each critical point. If the second derivative changes from negative to positive at a critical point, it indicates a local minimum; conversely, if it changes from positive to negative, there’s a local maximum. In the exercise, checking \( f''(x) \) around the critical point \( x = -\frac{2}{3} \), we see a shift from negative to positive, indicating that \( x = -\frac{2}{3} \) is a relative minimum for \( f'(x) \). However, at \( x = 0 \), no such sign change confirms a relative max or min, illustrating that not all critical points lead to relative extrema. Understanding relative extrema through calculus allows more sophisticated insights into the behavior and shape of the function's graph.