Problem 7 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

(a) Show that the derivative of \(\arccos x\) is \(\frac{-1}{\sqrt{1-x^{2}}}\). (b) What is the domain of \(\arccos x\) ? (c) What is the range of \(\arccos x\) ? (d) Where is the graph of \(\arccos x\) decreasing? (e) Where is the graph of \(\arccos x\) concave up? Concave down? If there is a point of in ection, where is it? (f) Graph \(f(x)=\arccos x\).

Step-by-Step Solution

Answer

(a) The derivative of \(arccos x\) is \(\frac{-1}{\sqrt{1-x^{2}}}\). (b) The domain of \(arccos x\) is \([-1,1]\). (c) The range of \(arccos x\) is \([0, \pi]\). (d) The graph of \(arccos x\) is decreasing on its entire domain. (e) The graph of \(arccos x\) is concave up on \(-1 < x < 0\) and concave down on \(0 < x < 1\). There are no points of inflection. (f) The graph of \(f(x) = arccos x\) starts at the point (1,0) and goes to the point (-1, \pi), it is always decreasing and is concave up on \(-1 < x < 0\) and concave down on \(0 < x < 1\).

Step 1: Find the derivative of \(\arccos x\)

By using chain rule, one arrives at \(y = \arccos x\), then \(x = \cos y\), and the derivative of \(\cos y\) is \(-\sin y\). By using the Pythagorean identity \(1 = \cos^{2}y + \sin^{2}y \), we solve for \(-\sin y\) to get \(\frac{-1}{\sqrt{1-x^{2}}}\).

Step 2: Find the domain of \(\arccos x\)

The domain of a function refers to all possible input values (x-values). For \(\arccos x\), this would be the set of all x such that \(-1 ≤ x ≤ 1.\)

Step 3: Find the range of \(\arccos x\)

The range of a function refers to all possible output values (y-values). For \(\arccos x\), the y-values span from \(0\) to \(\pi\). Therefore, the range is \(0 ≤ y ≤ \pi\).

Step 4: Determine where the graph of \(\arccos x\) is decreasing

The graph of \(\arccos x\) is decreasing on its entire domain of \(-1 ≤ x ≤ 1\). This is because its derivative is always negative.

Step 5: Determine the concavity and point of inflection of the graph of \(\arccos x\)

The second derivative of \(\arccos x\) is \(\frac{x} {(1-x^{2})^{(3/2)}}\). The function is undefined at \(x = -1\) and \(x = 1\), and does not change sign when going from less than -1 to greater than -1 or from less than 1 to greater than 1, therefore, there are no points of inflection. It is clear that when \(x < 0\), the second derivative is greater than 0, suggesting that it is concave up on \(-1 < x < 0\). Similarly, when \(x > 0\), the second derivative is less than 0, meaning it is concave down on \(0 < x < 1\).

Step 6: Graph \(f(x)=\arccos x\)

To graph \(f(x)=\arccos x\), take into account the findings above about the domain, range, decreasing nature and concavities. The graph starts at the point \((1,0)\) and goes to the point \((-1, \pi)\), it is always decreasing and is concave up on \(-1 < x < 0\) and concave down on \(0 < x < 1\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse the effect of regular trigonometric functions. The \( ext{arccosine}\) or \(\arccos x\), specifically, is the inverse of the cosine function. When you apply \(\arccos\), it converts a cosine value back to an angle.
For example, if \(\cos(\theta) = 0.5\), \(\arccos(0.5) = \frac{\pi}{3}\) because \(\cos(\frac{\pi}{3}) = 0.5\).
Inverse trigonometric functions have unique properties:

• The domain of \(\arccos x\) is from \(-1\) to \(1\), as these are the bounds within which cosine values fall.
• The range is from \(0\) to \(\pi\), which covers the possible output angles.

Due to these constraints, inverse functions have limited ranges but unique outputs for each input.

Chain Rule

The chain rule is a fundamental differentiation technique used when dealing with composite functions. It states that to differentiate a composite function \(f(g(x))\), you take the derivative of the outer function and multiply it by the derivative of the inner function.

When differentiating \(\arccos x\), consider reversing the roles:

• Start with \(y = \arccos x\), which implies \(x = \cos y\).
• Differentiate \(\cos y\) with respect to \(y\), which gives \(-\sin y\).
• By using the Pythagorean identity \(1 = \cos^2(y) + \sin^2(y)\), realize that \(-\sin y = \sqrt{1-x^2}\).

Combining these, the derivative of \(\arccos x\) is \(-\frac{1}{\sqrt{1-x^2}}\), showing how the chain rule intertwines these steps.

Function Domain and Range

Understanding the domain and range of \(\arccos x\) is crucial for working with the function.

**Domain**

• The domain refers to all possible input values for \(x\).
• For \(\arccos x\), valid inputs range from \(-1\) to \(1\). This is because cosine values only exist within this interval.

**Range**

• Denotes all potential outputs, or \(y\) values.
• For \(\arccos x\), possible outputs range from \(0\) to \(\pi\).

This means the angle output can vary between \(0\) and \(\pi\), covering a half-circle in radians.

Graphing Functions

Graphing \(\arccos x\) gives a visual representation of its behavior across its domain.

To plot \(f(x) = \arccos x\):

• The graph begins at the point \( (1,0)\) and extends to \((-1, \pi)\).
• The function is decreasing over the entire domain of \(-1 \leq x \leq 1\), as indicated by its negative derivative.
• It is concave up when \(-1 < x < 0\) and concave down when \(0 < x < 1\).

Understanding these properties allows you to draw the curve accurately, noting its symmetry and curvature changes.