Problem 3 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 1 through 10 , determine intervals of increase and decrease and intervals of concavity for the given function. Then sketch the graph of the function. Be sure to show all key features such as intercepts, asymptotes, high and low points, points of inflection, cusps, and vertical tangents. $$ f(x)=3 x^{3}-4 x^{2}-12 x+17 $$

Step-by-Step Solution

Answer

The function is increasing on \( -\infty, -\frac{2}{3}) \text{ and } \(\frac{14}{9}, \infty ) \ and decreasing on \( -\frac{2}{3}, \frac{14}{9})\. It is concave down on \( -\infty, \frac{4}{9}) \ and concave up on \( \frac{4}{9}, \infty )\. The y-intercept is 17.

Step 1: Find the first derivative

Differentiate the function to obtain the first derivative. \[ f'(x) = \frac{d}{dx}(3x^3 - 4x^2 - 12x + 17) = 9x^2 - 8x - 12 \]

Step 2: Determine critical points

Set the first derivative equal to zero and solve for \(x\): \[ 9x^2 - 8x - 12 = 0 \] Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 9\), \(b = -8\), and \(c = -12\): \[ x = \frac{8 \pm \sqrt{64 + 432}}{18} \] Simplify to get the critical points: \[ x = \frac{8 \pm 20}{18} \Rightarrow x = \frac{28}{18} = \frac{14}{9} \text{ and } x = \frac{-12}{18} = -\frac{2}{3} \]

Step 3: Determine intervals of increase and decrease

Test the first derivative in the intervals created by the critical points: 1. Choose a point in each interval and substitute into \(f'(x)\): - For \(x < -\frac{2}{3}\), choose \(x = -1\) - For \(-\frac{2}{3} < x < \frac{14}{9}\), choose \(x = 0\) - For \(x > \frac{14}{9}\), choose \(x = 2\)2. Determine the sign of \(f'(x)\) in each interval. - For \(f'(-1) = 9(-1)^2 - 8(-1) - 12 = 9 + 8 - 12 = 5 > 0\) - For \(f'(0) = 9(0)^2 - 8(0) - 12 = -12 < 0\) - For \(f'(2) = 9(2)^2 - 8(2) - 12 = 36 - 16 - 12 = 8 > 0\) Therefore, the function is increasing on \(( -\infty, -\frac{2}{3})\) and \((\frac{14}{9}, \infty)\); it is decreasing on \((-\frac{2}{3}, \frac{14}{9})\).

Step 4: Find the second derivative

Differentiate the first derivative to obtain the second derivative: \[ f''(x) = \frac{d}{dx}(9x^2 - 8x - 12) = 18x - 8 \]

Step 5: Determine intervals of concavity and points of inflection

Set the second derivative equal to zero to find points of inflection: \[ 18x - 8 = 0 \Rightarrow x = \frac{8}{18} = \frac{4}{9} \]Test the second derivative in intervals around \(x = \frac{4}{9}\):- For \(x < \frac{4}{9}\), choose \(x = 0\)- For \(x > \frac{4}{9}\), choose \(x = 1\)Determine the sign of \(f''(x)\) in each interval:- For \(f''(0) = 18(0) - 8 = -8 < 0 \Rightarrow \text{concave down on } (-\infty, \frac{4}{9})\)- For \(f''(1) = 18(1) - 8 = 10 > 0 \Rightarrow \text{concave up on } (\frac{4}{9}, \infty)\)Therefore, the point of inflection is at \(x = \frac{4}{9}\).

Step 6: Determine intercepts and other key features

Find the y-intercept by evaluating \(f(0)\): \[f(0) = 3(0)^3 - 4(0)^2 - 12(0) + 17 = 17\]There are no obvious vertical asymptotes as this is a polynomial. To find other intercepts, solve \(f(x) = 0\). This can be done through numerical or graphical methods as exact factorization is complex.

Step 7: Sketch the graph

Plot the key features including intercepts, critical points, and points of inflection. Using the determined intervals of increase/decrease and concavity to shape the curve correctly.

Key Concepts

derivatives

A derivative represents the rate at which a function changes at any given point. It's like a slope of the tangent line at a specific point on a curve. In our exercise, we have the function \(f(x) = 3x^3 - 4x^2 - 12x + 17\). To find how this function behaves, we need to compute its first and second derivatives.
The first derivative is obtained by differentiating the original function: \(f'(x) = 9x^2 - 8x - 12\). This first derivative helps us find the slope and thus tells us where the function is increasing or decreasing.
To dig deeper, we also compute the second derivative: \(f''(x) = 18x - 8\). This derivative informs us about the concavity of the function, helping to identify where the graph curves upwards or downwards.

critical points

Critical points are where the first derivative is zero or undefined. They are potential points where the function could have local maxima or minima. For our function, we set the first derivative equal to zero and solve: \[9x^2 - 8x - 12 = 0\]
Solving using the quadratic formula, we get \(x = \frac{14}{9}\) and \(x = -\frac{2}{3}\). These values divide the function into intervals and help us understand its behavior at specific points. By testing the sign of the first derivative in these intervals, we determine whether the function is increasing or decreasing around these critical points.

concavity

Concavity refers to the direction a curve bends. If a function is concave up, it looks like a cup and opens upwards. On the other hand, concave down looks like an upside-down cup. To determine concavity, we use the second derivative. For the given function, we find \(f''(x) = 18x - 8\).
We set this second derivative to zero to find potential inflection points: \[18x - 8 = 0 \rightarrow x = \frac{4}{9} \]
Testing values around \(x = \frac{4}{9}\), if \(f''(x) > 0\), the graph is concave up. If \(f''(x) < 0\), the graph is concave down. For instance, at \(x = 0\), \(f''(0) = -8\), indicating concave down. At \(x = 1\), \(f''(1) = 10\), indicating concave up.

intervals of increase and decrease

Intervals of increase occur where the first derivative \(f'(x) > 0\), and decrease where \(f'(x) < 0\). From the critical points found in our problem, we examine the sign of \(f'(x)\) in the intervals: \(( -\infty, -\frac{2}{3})\), \((-\frac{2}{3}, \frac{14}{9})\), and \((\frac{14}{9}, \infty)\).
By testing points within each interval: \[ \text{For} \, x < -\frac{2}{3}, \, f'(-1) = 5 \rightarrow \text{positive, meaning increasing} \]
\[ \text{For} \, -\frac{2}{3} < x < \frac{14}{9}, \, f'(0) = -12 \rightarrow \text{negative, meaning decreasing} \]
\[ \text{For} \, x > \frac{14}{9}, \, f'(2) = 8 \rightarrow \text{positive, meaning increasing} \]
Hence, the function increases on the intervals \(( -\infty, -\frac{2}{3})\) and \((\frac{14}{9}, \infty)\), and decreases on \((-\frac{2}{3}, \frac{14}{9})\).

points of inflection

Points of inflection are where the concavity of a function changes. These points show where the graph switches from concave up to concave down or vice versa. To find these points, we set the second derivative \(f''(x)\) to zero and solve: \[18x - 8 = 0 \rightarrow x = \frac{4}{9} \]
This \(x = \frac{4}{9}\) is a candidate for a point of inflection. By testing values around \(x = \frac{4}{9}\), we see how concavity changes: For \(x < \frac{4}{9}\) (e.g., \(x = 0\)), \(f''(0) = -8 \rightarrow \text{concave down}\). For \(x > \frac{4}{9}\) (e.g., \(x = 1\)), \(f''(1) = 10 \rightarrow \text{concave up}\). Hence, the function switches concavity at \(x = \frac{4}{9}\), confirming it as a point of inflection.