Question

In Problems 19-28, determine where the graph of the given function is increasing, decreasing, concave up, and concave down. Then sketch the graph (see Example 4). $$ f(x)=x^{3}-12 x+1 $$

Short answer

The graph is increasing on (-∞,-2) and (2,∞), decreasing on (-2,2); concave up on (0,∞), concave down on (-∞,0).

Step-by-step solution

Find the First Derivative

To analyze where the graph is increasing or decreasing, we find the first derivative of the function: \[ f'(x) = \frac{d}{dx}(x^3 - 12x + 1) = 3x^2 - 12. \]

Identify Critical Points

Set the first derivative equal to zero to find critical points:\[ 3x^2 - 12 = 0. \]Solving the equation, we get:\[ x^2 = 4 \Rightarrow x = \pm 2. \]

Determine Increasing and Decreasing Intervals

Using test points in intervals around the critical points (-∞, -2), (-2, 2), and (2, ∞), we determine the sign of the first derivative.- For \(x < -2\), choose \(x = -3\): \(f'(-3) = 27 - 12 = 15 > 0\) (increasing).- For \(-2 < x < 2\), choose \(x = 0\): \(f'(0) = -12 < 0\) (decreasing).- For \(x > 2\), choose \(x = 3\): \(f'(3) = 27 - 12 = 15 > 0\) (increasing).

Find the Second Derivative

To determine concavity, find the second derivative:\[ f''(x) = \frac{d}{dx}(3x^2 - 12) = 6x. \]

Identify Inflection Points

Set the second derivative equal to zero to find inflection points:\[ 6x = 0 \Rightarrow x = 0. \]

Determine Concavity Intervals

Using test points around the inflection point (x = 0), determine the sign of the second derivative.- For \(x < 0\), choose \(x = -1\): \(f''(-1) = -6 < 0\) (concave down).- For \(x > 0\), choose \(x = 1\): \(f''(1) = 6 > 0\) (concave up).

Sketch the Graph

Using the above information:- **Increasing** on \((-∞, -2) \cup (2, ∞)\);- **Decreasing** on \((-2, 2)\);- **Concave Down** on \((-∞, 0)\);- **Concave Up** on \((0, ∞)\);- **Points of interest** include critical points at \(x = -2, 2\) and inflection point at \(x = 0\). Sketch the graph considering these behaviors.

Key concepts

Graphing Functions

Graphing a function is the process of visually representing the equation on a coordinate plane. For the function given, \( f(x) = x^3 - 12x + 1 \), graphing involves understanding how this function behaves over different values of \( x \). By plotting several key points and understanding its overall shape, we can construct the entire graph.

• Start by identifying where the function intercepts the axes. The y-intercept occurs where \( x = 0 \), giving \( f(0) = 1 \).
• Calculate the x-intercepts by solving \( x^3 - 12x + 1 = 0 \), which may require numerical methods for exact results.
• Note the general shape dictated by the \( x^3 \) term, indicating the function's end-behavior: one end goes up, and the other goes down.

Once critical points and concavity information are determined, finishing the graph helps better visualize the intervals of increase and decrease. This visualization reinforces a deeper understanding of the function's dynamical behavior.

Derivatives

In calculus, a derivative represents how a function changes as its input changes. For any given function \( f(x) \), its first derivative \( f'(x) \) is the fundamental tool for analyzing this change.

• The first derivative \( f'(x) = 3x^2 - 12 \) signifies the slope of the tangent line at any point \( x \). Positive slope implies increasing values, while a negative slope indicates decreasing values.
• Using \( f'(x) \), we can pinpoint critical points where the function doesn't increase or decrease—a great tool for identifying turning points.

Differentiation simplifies complex functions into understandable chunks, enabling us to make predictions about the function's behavior across its domain.

Critical Points

Critical points are places on the graph where the derivative equals zero or is undefined. These points are crucial because they often signify changing directions of the function—from increasing to decreasing, or vice versa.

• For \( f(x) = x^3 - 12x + 1 \), setting the first derivative equal to zero \( 3x^2 - 12 = 0 \) helps locate these points. Solving gives us \( x = \pm2 \).
• At these critical points—\( x = -2 \) and \( x = 2 \)—the function potentially shifts from climbing to descending or vice versa, marking them as peaks or valleys.

Knowing critical points provides a map of significant changes in direction, which is crucial when sketching an accurate representation of the function's graph.

Concavity

Concavity describes the direction a curve opens and is determined by the second derivative \( f''(x) \). It helps differentiate between types of bends in the graph.

• The second derivative \( f''(x) = 6x \) shows how the slope changes and aids in identifying intervals where the graph bends upwards (concave up) or downwards (concave down).
• Setting \( f''(x) = 0 \) locates potential inflection points, where the concavity switches; for this function, it occurs at \( x = 0 \).
• By evaluating points around the inflection point, we discover that: to the left of \( x = 0 \), the function is concave down, and to the right, it is concave up.

Understanding concavity helps us not only with graph sketching but also with interpreting the nature of change—the extent of acceleration or deceleration—of the function.