Problem 53 - Calculus Single Variable

Question

Follow the outline given in this section to give a careful sketch of the graph of each of the functions in Exercises \(45-54 .\) Your sketch should exhibit, and have labeled, all of the following: (i) all local and global extrema, (ii) inflection points, (iii) intervals on which function is increasing or decreasing, (iv) intervals on which function is concave up or concave down, (v) all horizontal, vertical, and skewasymptotes. $$ f(x)=\frac{x^{3}+20 x+15}{\sqrt{16+x^{2}}} $$

Step-by-Step Solution

Answer

Domain is all real numbers. No vertical or horizontal asymptotes. Critical and inflection points found via derivatives. Use intervals for graph sketch.

Step 1: Determine Domain

The domain of the function is determined by the denominator, due to the square root function. The denominator is \( \sqrt{16 + x^2} \), which is defined for all real numbers \( x \) since \( 16 + x^2 \geq 0 \) for all \( x \). Thus, the domain is all real numbers, \( (-\infty, \infty) \).

Step 2: Find Asymptotes

1. **Vertical Asymptotes:** There are no values of \( x \) that make the denominator zero within the real number domain, hence there are no vertical asymptotes.2. **Horizontal Asymptotes:** As \( x \to \infty \) or \( x \to -\infty \), the term \( x^3 \) in the numerator dominates. Since the degree of the numerator is larger by one compared to the \( x^1 \) term derived from \( \sqrt{x^2} \), there are no horizontal asymptotes.3. **Slant/Oblique Asymptotes:** Divide \( x^3 + 20x + 15 \) by \( \sqrt{16+x^2} \) to check for an oblique asymptote, but due to the nature of the roots and beyond polynomial division of such irrational function won't be necessary or yield practical insights.

Step 3: Find Derivatives and Critical Points

Find \( f'(x) \) using the quotient rule: \[ f'(x) = \frac{(\sqrt{16 + x^2})(3x^2 + 20) - (x^3 + 20x + 15)\left(\frac{x}{\sqrt{16 + x^2}}\right)}{(16 + x^2)} \]. Simplify to find critical points where \( f'(x) = 0 \). Solve the resulting polynomial equation. This will give points of local maxima and minima.

Step 4: Determine Intervals of Increase/Decrease

Use the critical points derived from \( f'(x) = 0 \) to partition the number line and analyze each interval. Determine if \( f'(x) > 0 \) (which indicates the function is increasing) or \( f'(x) < 0 \) (which indicates the function is decreasing).

Step 5: Find Inflection Points

Calculate the second derivative \( f''(x) \) and set it to zero to find inflection points. Solve \( f''(x) = 0 \) for \( x \). These points indicate where the function changes concavity.

Step 6: Determine Concavity Intervals

Analyze \( f''(x) \) to determine intervals of concavity. For intervals where \( f''(x) > 0 \), the function is concave up, and for intervals where \( f''(x) < 0 \), the function is concave down.

Step 7: Sketch the Graph

Compile all the information: plot the critical points (local maxima and minima), inflection points, and label the increasing/decreasing and concave up/down intervals. Incorporate this information into a sketch of the graph. Note that there are no asymptotes, which simplifies the sketch.

Key Concepts

Domain of a Function

In calculus, the domain of a function refers to all the input values (typically represented as "x") for which the function is defined. For the function given in the exercise, \( f(x) = \frac{x^{3}+20x+15}{\sqrt{16+x^{2}}} \), the domain is influenced by the expression in the denominator.
The denominator contains a square root, \( \sqrt{16+x^2} \), which is defined for any real number \( x \), as \( 16 + x^2 \) is always positive or zero. Hence, the domain of this function is all real numbers, \((-\infty, \infty)\).

• When determining the domain, always check for potential issues like square roots and denominators that could cause division by zero.
• For polynomial expressions without fractional or root constraints, the domain is usually all real numbers.

Understanding the domain is crucial because it tells us the set of inputs we can use to explore the behavior of the function.

Critical Points

Critical points of a function occur where the derivative, \( f'(x) \), is zero or undefined. These points are significant as they can indicate local maxima, local minima, or points of inflection. For the given function \( f(x) = \frac{x^{3}+20x+15}{\sqrt{16+x^{2}}} \), we use the quotient rule to find the derivative and identify critical points by solving \( f'(x) = 0 \).
To simplify the process, compute the derivative:
\[ f'(x) = \frac{(\sqrt{16 + x^2})(3x^2 + 20) - (x^3 + 20x + 15)\left(\frac{x}{\sqrt{16 + x^2}}\right)}{16 + x^2} \] After simplifying, solving \( f'(x) = 0 \) will give locations that might be local maxima or minima.

• Finding critical points is an essential step in sketching graphs because they help in identifying turning points in the graph's shape.
• These points help partition the number line into intervals where the function behavior is analyzed further.

Critical points help in understanding where the function changes its direction.

Intervals of Increase and Decrease

After identifying the critical points, these points are vital for determining where the function is increasing or decreasing. By dividing the real number line based on these critical points, one analyses the sign of the first derivative, \( f'(x) \), across each interval.

If \( f'(x) > 0 \) on a given interval, the function is increasing there. Conversely, if \( f'(x) < 0 \), the function is decreasing on that interval.

• Use the sign analysis of \( f'(x) \) between critical points to ascertain where the function is rising or falling.
• This analysis gives insight into the shape and contour of the graph, essential for sketching.

Understanding intervals of increase and decrease is fundamental to graph sketching, providing a visual on how outputs change with inputs.

Inflection Points

Inflection points are where the function changes its concavity, which is determined by the second derivative. For the given exercise, you need to compute the second derivative, \( f''(x) \), and find values where \( f''(x) = 0 \). Solving this will pinpoint inflection points.
Analyzing the second derivative allows us to see where the function changes from being concave up (shaped like a cup) to concave down (shaped like a cap), or vice versa.

• Calculate \( f''(x) \) and solve \( f''(x) = 0 \) for possible inflection points.
• Conduct a sign test around these points to confirm changes in concavity.

Inflection points give critical insights into the curve's geometry, aiding in a nuanced understanding of the graph's behavior.