Problem 30 - Single Variable Calculus

Question

In Exercises \(5-38\), sketch the graph of the function using the curve- sketching guidelines on page \(348 .\) $$ f(x)=2 x-\tan x, \quad-\frac{\pi}{2}

Step-by-Step Solution

Answer

The graph of \(f(x) = 2x - \tan x\) on the interval \(-\frac{\pi}{2}

Step 1: Determine the Domain and Range

The given function is $$ f(x) = 2x - \tan x $$ Since only \(\tan x\) has restrictions within the given interval, the domain of the function will be: $$ -\frac{\pi}{2}

Step 2: Find x-intercepts and y-intercepts

To find the x-intercepts, we set \(f(x) = 0\): $$ 2x - \tan x = 0 $$ $$ 2x = \tan x $$ This equation is non-trivial to solve analytically, but we can use numerical methods. However, since the function is continuous on the given interval, there should be x-intercepts between the given interval. To find the y-intercept, we set \(x=0\): $$ f(0) = 2(0) - \tan(0) = 0 $$ Thus, the y-intercept is \((0,0)\).

Step 3: Identify Even or Odd Function

To determine if the function is even or odd, we check if it holds the following conditions: A function is even if \(f(x) = f(-x)\) and is odd if \(f(x) = -f(-x)\). Since $$ f(-x) = 2(-x) - \tan(-x) = -2x + \tan x \\ -f(-x) = 2x - \tan x, $$ the function is neither even nor odd.

Step 4: Determine Critical Points and Intervals of Increase and Decrease

Find the first derivative of the function to determine the critical points and intervals of increase and decrease: $$ f'(x) = 2 - \sec^2 x $$ Set the first derivative to zero and solve for \(x\) to find the critical points: $$ 2 - \sec^2 x = 0 $$ $$ \sec^2 x = 2 $$ $$ \cos^2 x = \frac{1}{2} $$ $$ \cos x = \pm\frac{1}{\sqrt{2}} $$ Within the interval \(-\frac{\pi}{2} 0\) for \(-\frac{\pi}{4}

Step 5: Identify Points of Inflection and Concavity Changes

Find the second derivative of the function to determine the points of inflection and concavity changes: $$ f''(x) = 0 - 2\sec^2x \tan x $$ This function doesn't have any points where \(f''(x) = 0\) within the given interval. Thus, there are no points of inflection.

Step 6: Sketch the Graph

With the information gathered, we can sketch the graph of the function \(f(x) = 2x - \tan x\): - Domain: \(-\frac{\pi}{2}

Key Concepts

Domain and Range of a Function

Understanding the domain and range of a function is like knowing the playing field for a game. The domain is where the function 'lives'—it's all the possible values you can plug into the function. For the function
\( f(x) = 2x - \tan x \),
the domain is restricted by the tangent function, since \(\tan x\) is undefined at \(\pm\frac{\pi}{2}\). Therefore, the function exists and is 'playable' between
\(-\frac{\pi}{2} The range, on the other hand, is the set of all possible outputs. For this function, since \(\tan x\) can take on any value from negative infinity to positive infinity, our game has no vertical limits — the range is \((-\f\infty, \f\infty)\).

X-intercepts and Y-intercepts

Finding where a function crosses the axes is like a treasure hunt. X-intercepts are where the function hits the horizontal axis. To find them for \(f(x) = 2x - \tan x\), you solve \(2x = \tan x\). This might require numerical methods, as the solution can be complex. But rest assured, there are x-intercepts in our domain.

The y-intercept is simpler: it's where the function crosses the vertical axis. In this case, at \(x=0\), the function simplifies to \(f(0) = 0\), meaning our treasure is at the origin (0,0).

Critical Points

Think of critical points as the peaks and valleys in a mountain range—they tell us where the function's slope changes from going up to going down, or vice versa. To find them for \(f(x) = 2x - \tan x\), we need to find where its derivative,
\(f'(x) = 2 - \sec^2 x\),
is zero or undefined. Solving \(2 - \sec^2 x = 0\) gives us the critical points at \(x_1 = -\frac{\pi}{4}\) and \(x_2 = \frac{\pi}{4}\). These critical points divide the domain into intervals where the function is either climbing up or sliding down.

Functions Increasing and Decreasing Intervals

A function’s mood can change: sometimes it’s rising (increasing), and sometimes it’s falling (decreasing). By checking our derivative \(f'(x) = 2 - \sec^2 x\),
we learn that:
- The function is in a 'down' mood, decreasing, for \(-\frac{\pi}{2} - It’s feeling 'up', increasing, for \(-\frac{\pi}{4} So, as we move along the graph, we can feel the function's mood swing from down to up and back down again as we pass through the critical points.

Concavity and Points of Inflection

Concavity tells us if the function is bending like a bowl (concave up) or like an arch (concave down). Points of inflection are like the twist in a roller coaster track—they're where the concavity changes. For our function, we look at the second derivative
\(f''(x) = 0 - 2\sec^2x \tan x\),
but find no points in our domain where it equals zero. This means there are no sudden twists—our function doesn't change its overall 'bending' direction within the domain. However, it’s always important to be attentive to the possibility of concavity changes when analyzing a function's behavior.

Previous

Problem 30

Next

Problem 30

Other exercises in this chapter

Problem 29

In Exercises \(25-40\), find the critical number \((s)\), if any, of the function. $$ f(x)=x^{3}-6 x+2 $$

View solution

Problem 30

Use Newton's method to obtain an approximation of the root of \(\cos ^{-1} x-x=0\) accurate to three decimal places.

View solution

Problem 30

Packaging A rectangular box is to have a square base and a volume of \(20 \mathrm{ft}^{3}\). If the material for the base costs \(\$ 0.30\) per square foot, the

View solution

Problem 30

Find the limit. $$ \text { 30. } \lim _{s \rightarrow-\infty}\left(\frac{s}{s+1}-\frac{s^{2}}{2 s^{2}+1}\right) $$

View solution

Key Concepts

Domain and Range of a Function

Understanding the domain and range of a function is like knowing the playing field for a game. The domain is where the function 'lives'—it's all the possible values you can plug into the function. For the function
\( f(x) = 2x - \tan x \),
the domain is restricted by the tangent function, since \(\tan x\) is undefined at \(\pm\frac{\pi}{2}\). Therefore, the function exists and is 'playable' between
\(-\frac{\pi}{2} The range, on the other hand, is the set of all possible outputs. For this function, since \(\tan x\) can take on any value from negative infinity to positive infinity, our game has no vertical limits — the range is \((-\f\infty, \f\infty)\).

X-intercepts and Y-intercepts

Finding where a function crosses the axes is like a treasure hunt. X-intercepts are where the function hits the horizontal axis. To find them for \(f(x) = 2x - \tan x\), you solve \(2x = \tan x\). This might require numerical methods, as the solution can be complex. But rest assured, there are x-intercepts in our domain.

The y-intercept is simpler: it's where the function crosses the vertical axis. In this case, at \(x=0\), the function simplifies to \(f(0) = 0\), meaning our treasure is at the origin (0,0).

Critical Points

Think of critical points as the peaks and valleys in a mountain range—they tell us where the function's slope changes from going up to going down, or vice versa. To find them for \(f(x) = 2x - \tan x\), we need to find where its derivative,
\(f'(x) = 2 - \sec^2 x\),
is zero or undefined. Solving \(2 - \sec^2 x = 0\) gives us the critical points at \(x_1 = -\frac{\pi}{4}\) and \(x_2 = \frac{\pi}{4}\). These critical points divide the domain into intervals where the function is either climbing up or sliding down.

Functions Increasing and Decreasing Intervals

A function’s mood can change: sometimes it’s rising (increasing), and sometimes it’s falling (decreasing). By checking our derivative \(f'(x) = 2 - \sec^2 x\),
we learn that:
- The function is in a 'down' mood, decreasing, for \(-\frac{\pi}{2} - It’s feeling 'up', increasing, for \(-\frac{\pi}{4} So, as we move along the graph, we can feel the function's mood swing from down to up and back down again as we pass through the critical points.

Concavity and Points of Inflection

Concavity tells us if the function is bending like a bowl (concave up) or like an arch (concave down). Points of inflection are like the twist in a roller coaster track—they're where the concavity changes. For our function, we look at the second derivative
\(f''(x) = 0 - 2\sec^2x \tan x\),
but find no points in our domain where it equals zero. This means there are no sudden twists—our function doesn't change its overall 'bending' direction within the domain. However, it’s always important to be attentive to the possibility of concavity changes when analyzing a function's behavior.