Problem 62
Question
a. Show that $$f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}$$ is an increasing function of \(x\) b. Show that $$g(x)=\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}$$ is a decreasing function of \(x\) c. Show that $$\frac{d t}{d x}=\frac{x}{c_{1} \sqrt{a^{2}+x^{2}}}-\frac{d-x}{c_{2} \sqrt{b^{2}+(d-x)^{2}}}$$ is an increasing function of \(x\)
Step-by-Step Solution
Verified Answer
a. \( f(x) \) is increasing.
b. \( g(x) \) is decreasing.
c. \( \frac{dt}{dx} \) is increasing.
1Step 1: Derivative of f(x)
To show that a function is increasing, we need to show its derivative is positive. Compute the derivative of \( f(x) = \frac{x}{\sqrt{a^{2}+x^{2}}} \) using the quotient rule. Let \( u = x \) and \( v = \sqrt{a^2 + x^2} \). Then \( u' = 1 \) and \( v' = \frac{x}{\sqrt{a^2 + x^2}} \). The derivative is:\[\frac{d}{dx} f(x) = \frac{v u' - u v'}{v^2} = \frac{\sqrt{a^2 + x^2} \cdot 1 - x \cdot \frac{x}{\sqrt{a^2 + x^2}}}{a^2 + x^2}\]Simplify the expression above to show:\[\frac{d}{dx} f(x) = \frac{a^2}{(a^2 + x^2)^{3/2}}\]Since \( a^2 > 0 \), the derivative is always positive for \( x eq 0 \), thus \( f(x) \) is an increasing function.
2Step 2: Derivative of g(x)
Compute the derivative of \( g(x) = \frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}} \) using the quotient rule. Let \( u = d-x \) and \( v = \sqrt{b^2 + (d-x)^2} \). Then \( u' = -1 \) and \( v' = \frac{-(d-x)}{\sqrt{b^2 + (d-x)^2}} \). The derivative is:\[\frac{d}{dx} g(x) = \frac{v u' - u v'}{v^2} = \frac{\sqrt{b^2 + (d-x)^2} (-1) - (d-x) \left(\frac{-(d-x)}{\sqrt{b^2 + (d-x)^2}}\right)}{b^2 + (d-x)^2}\]Simplify the above to show:\[\frac{d}{dx} g(x) = -\frac{b^2}{(b^2 + (d-x)^2)^{3/2}}\]The expression \(-\frac{b^2}{(b^2 + (d-x)^2)^{3/2}}\) is negative for all \( x \), proving that \( g(x) \) is a decreasing function.
3Step 3: Determining the derivative of h(x)
The function is \( \frac{d t}{d x} = \frac{x}{c_{1} \sqrt{a^{2} + x^{2}}} - \frac{d-x}{c_{2} \sqrt{b^{2} + (d-x)^{2}}} \). To determine if it's increasing, consider the expression:\[h(x) = \frac{x}{c_{1} \sqrt{a^{2} + x^{2}}} + \frac{x-d}{c_{2} \sqrt{b^{2} + (d-x)^{2}}}\]Taking the derivative we find two prior results are used:\[\text{Derivative of } \frac{x}{\sqrt{a^{2} + x^{2}}} = \frac{a^2/c_1}{(a^2 + x^2)^{3/2}}\]\[\text{Derivative of } \frac{d-x}{\sqrt{b^{2} + (d-x)^{2}}} = -\frac{b^2/c_2}{(b^2 + (d-x)^2)^{3/2}}\]These derivatives are positive and negative, respectfully. Combining yields a positive expression and \( \frac{d t}{d x} \) is increasing.
Key Concepts
Understanding DerivativesWhat Makes A Function Increasing?Understanding Decreasing Functions
Understanding Derivatives
A derivative is a fundamental concept in calculus, representing the rate at which a function changes at any point. Imagine you're driving a car and you want to know how your speed changes from one moment to the next. The derivative acts like the speedometer, showing you the rate of change of distance over time.
In mathematical terms, the derivative of a function \( f(x) \) is denoted as \( f'(x) \) or \( \frac{d}{dx}f(x) \). It's calculated using different rules depending on the function involved. One such rule is the **quotient rule**, used when dealing with a ratio of functions. The quotient rule states that for functions \( u(x) \) and \( v(x) \), the derivative \( \frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} \).
Derivatives help us determine whether a function is increasing or decreasing, by looking at whether the derivative is positive or negative at a point or over an interval.
In mathematical terms, the derivative of a function \( f(x) \) is denoted as \( f'(x) \) or \( \frac{d}{dx}f(x) \). It's calculated using different rules depending on the function involved. One such rule is the **quotient rule**, used when dealing with a ratio of functions. The quotient rule states that for functions \( u(x) \) and \( v(x) \), the derivative \( \frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} \).
Derivatives help us determine whether a function is increasing or decreasing, by looking at whether the derivative is positive or negative at a point or over an interval.
What Makes A Function Increasing?
A function is said to be increasing if its derivative is positive over its domain. This means, as \( x \) increases, \( f(x) \) also increases. It's like walking uphill – each step you take, you go higher.
Consider the function \( f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}} \). To determine if \( f(x) \) is increasing, we compute its derivative and check its sign. Using the quotient rule, the derivative is \( \frac{a^2}{(a^2 + x^2)^{3/2}} \), which is positive because \( a^2 > 0 \) and the denominator is always positive for real numbers. As a result, since the derivative is positive for all \( x eq 0 \), \( f(x) \) is an increasing function.
Key points to remember:
Consider the function \( f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}} \). To determine if \( f(x) \) is increasing, we compute its derivative and check its sign. Using the quotient rule, the derivative is \( \frac{a^2}{(a^2 + x^2)^{3/2}} \), which is positive because \( a^2 > 0 \) and the denominator is always positive for real numbers. As a result, since the derivative is positive for all \( x eq 0 \), \( f(x) \) is an increasing function.
Key points to remember:
- A function is increasing when \( f'(x) > 0 \) for all \( x \) in its domain.
- Positive derivatives indicate upward slopes on a graph.
- The function grows as \( x \) increases.
Understanding Decreasing Functions
A function is classified as decreasing if its derivative is negative over its domain. In simple terms, as \( x \) moves ahead on the number line, the function value \( f(x) \) moves downward, like walking downhill.
Take for example the function \( g(x)=\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}} \). To see if it’s decreasing, we look at its derivative: \( -\frac{b^2}{(b^2 + (d-x)^2)^{3/2}} \). This derivative is negative because \( -b^2 < 0 \), ensuring that no matter the value of \( x \), \( g(x) \) slopes downwards.
Remember these essential points about decreasing functions:
Take for example the function \( g(x)=\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}} \). To see if it’s decreasing, we look at its derivative: \( -\frac{b^2}{(b^2 + (d-x)^2)^{3/2}} \). This derivative is negative because \( -b^2 < 0 \), ensuring that no matter the value of \( x \), \( g(x) \) slopes downwards.
Remember these essential points about decreasing functions:
- A function is decreasing when \( f'(x) < 0 \) universally for its domain.
- Negative derivatives indicate a downward slope of the graph.
- The function diminishes as \( x \) augments.
Other exercises in this chapter
Problem 61
Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x \sqrt{4-x^{2}}$$
View solution Problem 62
Verify the formulas by differentiation. $$\int \frac{1}{(x+1)^{2}} d x=\frac{x}{x+1}+C$$
View solution Problem 62
Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x^{2} \sqrt{3-x}$$
View solution Problem 62
Show that the functions in Exercises 61 and 62 have local extreme values at the given values of \(\theta\), and say which kind of local extreme the function has
View solution