Problem 72
Question
If \(f(x)=x^{m}, m\) being a non-negative integer, then the value of \(m\) for which \(f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)\), for all \(\alpha, \beta>0\), is (A) 1 (B) 2 (C) 0 (D) None of these
Step-by-Step Solution
Verified Answer
The value of \(m\) is 1.
1Step 1: Find the derivative
To find the derivative of the function \(f(x) = x^m\), we use the power rule for differentiation. The derivative is \(f'(x) = mx^{m-1}\).
2Step 2: Apply the condition
We need to find the integer \(m\) such that \(f'(\alpha + \beta) = f'(\alpha) + f'(\beta)\) for all \(\alpha, \beta > 0\). Substitute the derivative into the condition to get \(m(\alpha + \beta)^{m-1} = m\alpha^{m-1} + m\beta^{m-1}\).
3Step 3: Simplify the equation
Cancel out the common factor \(m\) from both sides (assuming \(m eq 0\)) to get \((\alpha + \beta)^{m-1} = \alpha^{m-1} + \beta^{m-1}\). This equation must hold true for all \(\alpha, \beta > 0\).
4Step 4: Analyze possible values of \(m\)
For \((\alpha + \beta)^{m-1} = \alpha^{m-1} + \beta^{m-1}\) to hold for all positive \(\alpha\) and \(\beta\), try simple cases. For \(m = 1\), it becomes \(\alpha + \beta = \alpha + \beta\), which is true. For \(m > 1\), the equation doesn't generally hold. If \(m = 0\), it simplifies to \(1 = 1 + 1\), which is not valid.
Key Concepts
Power Rule for DifferentiationDerivative FunctionProblem Solving in Calculus
Power Rule for Differentiation
The power rule is a quick way to find the derivative of a function of the form \( f(x) = x^m \), where \( m \) is a constant.
It states that if you have a function \( x^m \), the derivative is \( f'(x) = mx^{m-1} \).
This rule simplifies the process of finding the derivative significantly.
Here's why it works:
Remember, the power rule applies as long as \( m \) is any real number.
This elegantly transitions us into considering how this derivative is used in calculus problems, like the one in our exercise.
It states that if you have a function \( x^m \), the derivative is \( f'(x) = mx^{m-1} \).
This rule simplifies the process of finding the derivative significantly.
Here's why it works:
- You multiply the exponent \( m \) by the base \( x \).
- You then reduce the exponent by one, turning \( x^m \) into \( x^{m-1} \).
Remember, the power rule applies as long as \( m \) is any real number.
This elegantly transitions us into considering how this derivative is used in calculus problems, like the one in our exercise.
Derivative Function
The derivative function represents the rate at which a function's output value changes as its input value changes.
For our function, \( f(x) = x^m \), the derivative is given by \( f'(x) = mx^{m-1} \).
This derivative tells you how steep the curve of the function is at any point.
Understanding the idea of derivatives is crucial in calculus. It allows:
What this tells us is that we want the rate of change when adding two inputs to equal the sum of their individual rates of change.
This property greatly narrows down possible values for \( m \), helping solve the problem.
For our function, \( f(x) = x^m \), the derivative is given by \( f'(x) = mx^{m-1} \).
This derivative tells you how steep the curve of the function is at any point.
Understanding the idea of derivatives is crucial in calculus. It allows:
- Understanding instantaneous rates of change (like velocity in physics).
- Finding slopes of tangent lines to curves.
What this tells us is that we want the rate of change when adding two inputs to equal the sum of their individual rates of change.
This property greatly narrows down possible values for \( m \), helping solve the problem.
Problem Solving in Calculus
Solving calculus problems often requires applying known rules to find unknowns and verify conditions.
In the exercise provided, after finding the derivative using the power rule, we apply the condition \( f'(\alpha + \beta) = f'(\alpha) + f'(\beta) \).
Here's a breakdown of the process:
In this case, \( m = 1 \) works because \( \alpha + \beta = \alpha + \beta \) simplifies perfectly.
With \( m = 0 \) or \( m > 1 \), the condition doesn't hold, informing our answer choice.
This problem exemplifies the systematic approach often needed in calculus to find solutions while using derivative functions.
In the exercise provided, after finding the derivative using the power rule, we apply the condition \( f'(\alpha + \beta) = f'(\alpha) + f'(\beta) \).
Here's a breakdown of the process:
- Use the derivative obtained from the power rule: \( f'(x) = mx^{m-1} \).
- Substitute this into the condition: \( m(\alpha + \beta)^{m-1} = m\alpha^{m-1} + m\beta^{m-1} \).
- Simplify by canceling the \( m \) (provided \( m eq 0 \)) to see if it holds for all positive \( \alpha \) and \( \beta \).
In this case, \( m = 1 \) works because \( \alpha + \beta = \alpha + \beta \) simplifies perfectly.
With \( m = 0 \) or \( m > 1 \), the condition doesn't hold, informing our answer choice.
This problem exemplifies the systematic approach often needed in calculus to find solutions while using derivative functions.
Other exercises in this chapter
Problem 70
If \(y^{3}-y=2 x\), then \(\left(x^{2}-\frac{1}{27}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=\) (A) \(y\) (B) \(\frac{y}{3}\) (C) \(\frac{y}{9}\) (D) \(
View solution Problem 71
If \(x
View solution Problem 74
If \(f(x-y), f(x) \cdot f(y)\) and \(f(x+y)\) are in A.P. for all \(x, y\) and \(f(0) \neq 0\), then (A) \(f(2)=f(-2)\) (B) \(f(3)+f(-3)=0\) (C) \(f^{\prime}(2)
View solution Problem 77
A function \(f:(0, \infty) \rightarrow R\) satisfies the equation \(f(x y)=2 f(x)-f\left(\frac{x}{y}\right)\) If \(f\) is differentiable on \(R\) and \(f(1)=0,
View solution