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Chapter 5: Problem 17

If \(f(x)=m x^{2}+n x+p\), Then \(f^{\prime}(1)+f^{\prime}(4)-f^{\prime}(5)\) is equal to [MPPET-2008] (a) \(m\) (b) \(-m\) (c) \(n\) (d) \(-n\)

Short Answer

Expert verified
The expression evaluates to 0, which doesn't match any provided options.

Step by step solution

01

Identify the Problem

We need to compute the expression \(f^{\prime}(1) + f^{\prime}(4) - f^{\prime}(5)\), where \(f(x) = mx^2 + nx + p\). To proceed, we first need \(f'(x)\), the derivative of \(f(x)\).
02

Find the Derivative

Compute the derivative \(f'(x)\) of the function \(f(x) = mx^2 + nx + p\). Using standard derivative rules, the derivative is \(f'(x) = 2mx + n\).
03

Compute Specific Derivatives

Substitute the values 1, 4, and 5 into \(f'(x) = 2mx + n\) to find \(f'(1)\), \(f'(4)\), and \(f'(5)\). - \(f'(1) = 2m(1) + n = 2m + n\)- \(f'(4) = 2m(4) + n = 8m + n\)- \(f'(5) = 2m(5) + n = 10m + n\)
04

Substitute and Simplify

Substitute the computed values into the expression \(f'(1) + f'(4) - f'(5)\):\((2m + n) + (8m + n) - (10m + n)\).Simplify this to: \(2m + n + 8m + n - 10m - n = 0m + 0n = 0\).
05

Conclude

The result of the expression \(f'(1) + f'(4) - f'(5)\) is 0, meaning the expression evaluates to zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Functions
Quadratic functions are a type of polynomial function specifically of degree two. They have the general form:
  • \(f(x) = ax^2 + bx + c\)
where:
  • \(a, b,\) and \(c\) are constants,
  • \(a eq 0\) (to ensure it is a quadratic, not linear),
  • \(x\) represents the variable.
The graph of a quadratic function is a parabola, which can open upwards if \(a > 0\) or downwards if \(a < 0\). The shape and position of the parabola are determined by the coefficients \(a\), \(b\), and \(c\). Quadratic functions are widely used in fields like physics, engineering, finance, and many other disciplines where relationships can be modeled by parabolas.
Derivative Rules
In calculus, derivatives are used to measure how a function changes as its input changes. They are fundamental in understanding motion and change. For a quadratic function such as \( f(x) = mx^2 + nx + p \), the process to find the derivative involves several rules:
  • The Power Rule: For any term \(x^n\), the derivative is \(nx^{n-1}\).
  • The Constant Rule: The derivative of any constant is zero.
  • The Sum Rule: The derivative of a sum is the sum of the derivatives.
Using these rules, the derivative of the function \( f(x) = mx^2 + nx + p \) is computed as follows:
  • \( f'(x) = \frac{d}{dx}(mx^2) + \frac{d}{dx}(nx) + \frac{d}{dx}(p) = 2mx + n \).
Understanding how these rules apply to derivatives makes it simpler to tackle complex problems involving rates of change.
Function Evaluation
Function evaluation involves substituting specific values into the function to determine the result. It's a critical step in calculus and mathematics in general. For a derivative \( f'(x) = 2mx + n \), evaluating it at specific points gives insight into the behavior of the original quadratic function at those points.
  • To find \( f'(1) \), substitute \( x = 1 \) into the derivative: \( f'(1) = 2m(1) + n = 2m + n \).
  • To find \( f'(4) \), substitute \( x = 4 \): \( f'(4) = 2m(4) + n = 8m + n \).
  • To find \( f'(5) \), substitute \( x = 5 \): \( f'(5) = 10m + n \).
These evaluations help us understand the rate of change of the function \( f(x) \) at specific input values, crucial for solving real-world problems.
Expression Simplification
Simplification in mathematics involves reducing expressions to their simplest form to make them easier to work with. Simplifying the expression \( f'(1) + f'(4) - f'(5) \) requires basic arithmetic operations:
  • Start with the values: \((2m + n) + (8m + n) - (10m + n)\).
  • Combine like terms: \(2m + 8m - 10m + n + n - n \).
  • Simplify: \(0m + 0n = 0\).
Simplifying expressions like this is crucial for solving mathematical problems. It eliminates unnecessary complexity, allowing you to focus on the solution without distraction. This exercise is a perfect illustration of how simplification can lead to realizing that the expression evaluates to zero.

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Most popular questions from this chapter

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