Problem 49
Question
Let \(f(x)=\sqrt{x-1}+\sqrt{x+24-10 \sqrt{x-1}} ; 1
Step-by-Step Solution
Verified Answer
Option (A): 0.
1Step 1: Simplify the Function
The given function is \( f(x) = \sqrt{x-1} + \sqrt{x+24-10\sqrt{x-1}} \). Let's simplify the second square root term. Set \( y = \sqrt{x-1} \), then \( x = y^2 + 1 \). Substitute this into the second term: \[ \sqrt{x+24-10\sqrt{x-1}} = \sqrt{y^2 + 1 + 24 - 10y} = \sqrt{y^2 - 10y + 25} = \sqrt{(y-5)^2} = |y-5| \] Since \( y = \sqrt{x-1} = \sqrt{y^2} \), and \( y = \sqrt{x-1} > 0 \) for \( x > 1 \), we find \( y - 5 \) is negative for \( 1 < x < 26 \) because \( y = \sqrt{x-1} < 5 \). Thus, \(|y-5| = -(y-5) = 5 - y\). So: \( \sqrt{x+24-10\sqrt{x-1}} = 5 - \sqrt{x-1} \). The function simplifies to: \[ f(x) = \sqrt{x-1} + (5 - \sqrt{x-1}) = 5 \] This shows that \( f(x) \) simplifies to a constant, 5.
2Step 2: Differentiate the Simplified Function
Since \( f(x) = 5 \) is a constant function, its derivative is 0. Therefore, \[ f^{\prime}(x) = 0 \] for all \( 1 < x < 26 \).
Key Concepts
Function SimplificationSquare RootsConstant FunctionDerivative
Function Simplification
Function simplification is essential for solving complex mathematical problems. In this exercise, the given function is \( f(x) = \sqrt{x-1} + \sqrt{x+24-10\sqrt{x-1}} \). After analyzing the expression, the exercise guides you to simplify it using substitution.
By letting \( y = \sqrt{x-1} \), we redefine the function in terms of \( y \). This allows us to rewrite the second square root in simpler terms. We'll see that the expression inside the square root \( x+24-10\sqrt{x-1} \) simplifies to \((y-5)^2\).
Simplifying further, we use the property of absolute values. The function then reduces to a constant value as \( f(x) = 5 \). Simplification turns complex combinations into more manageable forms. It is a powerful technique, often the first step in evaluating or differentiating intricate functions.
By letting \( y = \sqrt{x-1} \), we redefine the function in terms of \( y \). This allows us to rewrite the second square root in simpler terms. We'll see that the expression inside the square root \( x+24-10\sqrt{x-1} \) simplifies to \((y-5)^2\).
Simplifying further, we use the property of absolute values. The function then reduces to a constant value as \( f(x) = 5 \). Simplification turns complex combinations into more manageable forms. It is a powerful technique, often the first step in evaluating or differentiating intricate functions.
Square Roots
Square roots appear extensively across math problems, often requiring careful attention. In the original function, both terms include a square root expression. The transformation \( \sqrt{x-1} \) plays a significant role in simplifying the problem.
By setting \( \sqrt{x-1} = y \), we convert the variable \( x \) into a simpler form involving \( y \). This substitution focuses on the expression to unravel hidden relationships that ease calculations.
Exploring square roots, paired with basic algebraic identities, can uncover simpler expressions, easing the pathway toward differentiation or other calculus operations.
By setting \( \sqrt{x-1} = y \), we convert the variable \( x \) into a simpler form involving \( y \). This substitution focuses on the expression to unravel hidden relationships that ease calculations.
- The substitution transforms \( \sqrt{x+24-10\sqrt{x-1}} \) to \( |y-5| \), which leads to further simplification.
- As we continue, acknowledging the properties of square roots, such as the conditions \( \sqrt{x} \geq 0 \), ensures our solutions remain valid.
Exploring square roots, paired with basic algebraic identities, can uncover simpler expressions, easing the pathway toward differentiation or other calculus operations.
Constant Function
A constant function remains unchanged across its domain. After simplifying, the original exercise function \( f(x) \) turns into a constant function, specifically \( f(x) = 5 \).
This remarkable simplification indicates that the function value does not depend on \( x \) for the specified interval \( 1 < x < 26 \). To recognize and prove a function as constant is critical, particularly before proceeding to the differentiation stage. Here are properties that help identify constant functions:
Understanding that a function is constant can save calculation time. It immediately leads us to conclude that the derivative of this function will be zero.
This remarkable simplification indicates that the function value does not depend on \( x \) for the specified interval \( 1 < x < 26 \). To recognize and prove a function as constant is critical, particularly before proceeding to the differentiation stage. Here are properties that help identify constant functions:
- Ineffects from changes in \( x \) do not alter the function value.
- Graphically represented as a horizontal line on the Cartesian plane.
Understanding that a function is constant can save calculation time. It immediately leads us to conclude that the derivative of this function will be zero.
Derivative
Differentiating a function involves finding its rate of change with respect to the variable. The core derivative concept is finding the slope of the tangent line to the curve at any given point. In this exercise's case, \( f(x) \), upon being simplified to a constant, demonstrates the concept.
When a function like \( f(x) = 5 \) is constant, its derivative \( f'(x) \) is zero. The derivative of a constant \( c \) is zero because:
Recognizing such situations simplifies the differentiation process. It underlines the importance of confirming a function's nature before executing any calculated differentiation. In summary, familiarizing yourself with derivative concepts helps effortlessly tackle such problems.
When a function like \( f(x) = 5 \) is constant, its derivative \( f'(x) \) is zero. The derivative of a constant \( c \) is zero because:
- There is no change in \( y \)-values as \( x \) changes, hence no slope.
- The derivative of \( c \) with respect to any variable \( x \) does not vary.
Recognizing such situations simplifies the differentiation process. It underlines the importance of confirming a function's nature before executing any calculated differentiation. In summary, familiarizing yourself with derivative concepts helps effortlessly tackle such problems.
Other exercises in this chapter
Problem 46
If \(\sqrt{x+y}+\sqrt{y-x}=c\) then \(\frac{d^{2} y}{d x^{2}}\) equals (A) \(\frac{2}{c^{2}}\) (B) \(\frac{-2}{c^{2}}\) (C) \(\frac{2}{c}\) (D) \(\frac{-2}{c}\)
View solution Problem 47
Let \(f(x)=\prod_{k=1}^{n}(\cos (2 k-1) x+i \sin (2 k-1) x)\), then \((\operatorname{Re} f(x))^{\prime \prime}+i(\operatorname{Im} f(x))^{\prime \prime}\) is eq
View solution Problem 50
If \(f(x)=|x-2|\) and \(g(x)=f\\{f(x)\\}\), then \(g^{\prime}(x)\) for \(x>2\) is (A) \(-1\) (B) 1 (C) 0 (D) does not exist
View solution Problem 51
The derivative of the function represented parametrically as \(x=2 t-|t|, y=t^{3}+t^{2}|t|\) at \(t=0\) is (A) 0 (B) 1 (C) \(-1\) (D) does not exist
View solution