/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 The graph of \(f(x)=\frac{a x+b}... [FREE SOLUTION] | 91Ó°ÊÓ

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The graph of \(f(x)=\frac{a x+b}{(x-1)(x-4)}\) has a horizontal tangent line at (2,-1) Find \(a\) and \(b\). Check using a graphing calculator.

Short Answer

Expert verified
The values are \( a = 2 \) and \( b = 0 \).

Step by step solution

01

Understand the Function's Derivative

To find where the graph has a horizontal tangent, we need to find where the slope, or the first derivative, equals zero. Take the derivative of the function \( f(x) = \frac{ax+b}{(x-1)(x-4)} \) using the quotient rule. The derivative is \( f'(x) = \frac{d}{dx} \left( \frac{ax+b}{(x-1)(x-4)} \right) \).
02

Apply the Quotient Rule

The quotient rule states \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \) where \( u = ax + b \) and \( v = (x-1)(x-4) \). Compute \( u' = a \) and \( v' = 2x - 5 \). This gives the derivative: \[ f'(x) = \frac{(ax+b)(2x-5) - a(x-1)(x-4)}{((x-1)(x-4))^2} \].
03

Solve for Horizontal Tangent

A horizontal tangent occurs where \( f'(x) = 0 \). Therefore, solve \((ax+b)(2x-5) - a(x-1)(x-4) = 0 \) at the given point \( x = 2 \).
04

Substitute and Simplify

Substitute \( x = 2 \) into the equation \( (ax+b)(2x-5) - a(x-1)(x-4) = 0 \). Simplify: \[(2a + b)(4 - 5) - a(1)(-2) = 0 \Rightarrow -2a - b + 2a = 0 \].
05

Use Given Point to Formulate Equations

The point (2,-1) is on the curve, so substitute \( x = 2 \) and \( f(x) = -1\) into the original equation to form: \[ -1 = \frac{2a + b}{2(-2)} \Rightarrow 2a + b = 4 \].
06

Solve for a and b

From Step 4 and Step 5, we have two equations, \(-b = 0\) and \(2a + b = 4\). Solving these gives \( b = 0 \) and \( 2a = 4 \) thus \( a = 2 \).
07

Verify with Graphing Calculator

Graph the function with \( a = 2 \) and \( b = 0 \) as \( f(x) = \frac{2x}{(x-1)(x-4)} \) and check the slope approaches zero at the point (2,-1).

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Key Concepts

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

Quotient Rule
The Quotient Rule is a key tool in calculus, especially when dealing with the derivatives of functions that are expressed as a fraction or a quotient of two other functions. When you have a function of the form \( \frac{u}{v} \), the quotient rule helps you find its derivative, represented as \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \). Here:
  • \( u \) is the numerator function.
  • \( v \) is the denominator function.
  • \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.
In the problem at hand, the function \( f(x) = \frac{ax+b}{(x-1)(x-4)} \) requires the quotient rule to find its derivative. By setting \( u = ax + b \) and \( v = (x-1)(x-4) \), and knowing their derivatives \( u' = a \) and \( v' = 2x - 5 \), the rule allows us to construct the complex derivative needed for further calculations. This is essential in solving most calculus problems that involve divisions.
Horizontal Tangent
A horizontal tangent line is where the slope of a function is zero, which visually means the graph is flat at that point. In calculus, this corresponds to finding points where the first derivative of a function equals zero. In the problem we are working with, we are told that the graph of the function \( f(x) = \frac{ax+b}{(x-1)(x-4)} \) has a horizontal tangent at the point \((2, -1)\).
To find such a point, one solves for \( f'(x) = 0 \). This means setting the numerator of the derivative (after applying the quotient rule) equal to zero, as only then does the entire expression simplify to zero (since the denominator cannot be zero for continuity).
Horizontal tangents provide critical insights into the behavior of functions, indicating local maxima, minima, or inflection points, which are key aspects of the graph's shape and trend.
Graph of a Function
The graph of a function is a visual representation of all its values. For rational functions like \( f(x) = \frac{ax+b}{(x-1)(x-4)} \), understanding their graph involves identifying asymptotes, intercepts, and symmetries alongside horizontal tangents.
In our exercise, the graph has a horizontal tangent at \((2, -1)\), which means if we plot \( f(x) \) and examine it, we'll observe a flat line at this point. This direct observation confirms our calculations. One can see how variable coefficients \( a \) and \( b \) modify this graph’s shape and behavior, demonstrating the importance of parametric understanding.
  • Asymptotes occur where the denominator equals zero.
  • Intercepts occur where the numerator equals zero when the denominator is not zero.
Using a graphing calculator is often suggested to verify function behaviors visually, providing an intuitive grasp of analytically derived results.
Calculus Problem Solving
Solving calculus problems usually follows a structured approach, especially involving derivatives and function analysis. The exercise is a classic illustration of calculus concepts involving the calculation of derivatives using rules and understanding function behaviors.
  • Identify what the function looks like analytically and its domain.
  • Apply calculus techniques, like the quotient rule, to uncover properties such as derivative and slope.
  • Find specific function attributes, like horizontal tangents, optimally solving equations derived from derivative conditions.
  • Validate your findings using tools like graphing calculators, which compare your analytic result against a plotted curve.
By practicing structured problem-solving, such as employing derivative methods followed by graphical interpretation, calculus problems become more manageable, leading to effective analysis and comprehension of function behavior and characteristics.

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