Chapter 4: Problem 38
Find values of \(a\) and \(b\) so that the function \(f(x)=\) \(x^{2}+a x+b\) has a local minimum at the point (6,-5)
Short Answer
Expert verified
The values are \(a = -12\) and \(b = 31\).
Step by step solution
01
Understand the Problem
We need to find the values of \(a\) and \(b\) for the quadratic function \(f(x)=x^2 + ax + b\) such that the function has a local minimum at the point \((6,-5)\). This means at \(x = 6\), \(f(x) = -5\) and the derivative \(f'(x)\) should be 0.
02
Apply Derivative Condition
For the function to have a local minimum at \(x=6\), the derivative \(f'(x)\) should be zero at \(x=6\). The derivative of \(f(x)\) with respect to \(x\) is \(f'(x) = 2x + a\). Setting \(f'(6) = 0\), we have:\[ f'(6) = 2(6) + a = 0 \]\[ 12 + a = 0 \]Thus, we find \(a = -12\).
03
Apply Function Value Condition
We also know that the function value at \(x = 6\) is \(-5\), i.e., \(f(6) = -5\). Substituting \(a = -12\) into the function, we have:\[ f(6) = 6^2 + (-12)(6) + b = -5 \]\[ 36 - 72 + b = -5 \]\[ -36 + b = -5 \]Solving for \(b\), we get:\[ b = -5 + 36 = 31 \].
04
Verify the Conditions
Now that we have \(a = -12\) and \(b = 31\), let's verify the conditions:- The derivative at \(x=6\) is \(f'(6) = 12 - 12 = 0\) (Condition of local minimum is satisfied).- The function value at \(x=6\) is \(f(6)= 36 - 72 + 31 = -5\) (Matches the given point).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Derivatives in Quadratic Functions
Derivatives are a fundamental concept in calculus, used to determine the slope of a function at any point. For quadratic functions like \(f(x) = x^2 + ax + b\), the derivative helps us find where the slope is zero, which is crucial for identifying points like local minima or maxima.
- A derivative of a function gives us another function, indicating the rate of change of the original function. - When the derivative \(f'(x)\) equals zero, it means the slope of the tangent line to the curve is horizontal at that point, indicating a potential minimum or maximum.
In the context of our exercise, the derivative \(f'(x) = 2x + a\) must be zero at \(x = 6\) for a local minimum. By solving \(f'(6) = 0\), we learned that \(a = -12\). This simplicity in calculating derivatives is one of the reasons why quadratic functions are so approachable and useful in various mathematical applications.
- A derivative of a function gives us another function, indicating the rate of change of the original function. - When the derivative \(f'(x)\) equals zero, it means the slope of the tangent line to the curve is horizontal at that point, indicating a potential minimum or maximum.
In the context of our exercise, the derivative \(f'(x) = 2x + a\) must be zero at \(x = 6\) for a local minimum. By solving \(f'(6) = 0\), we learned that \(a = -12\). This simplicity in calculating derivatives is one of the reasons why quadratic functions are so approachable and useful in various mathematical applications.
Local Minimum and Its Characteristics
A local minimum refers to a point on a function where the value is less than or equal to the values at nearby points. In simpler terms, it's like being at the lowest point in a nearby region.
For a function to have a local minimum, the derivative at that point must be zero \(f'(x) = 0\). This zero slope denotes a flat tangent.- In addition to \(f'(x) = 0\), the second derivative \(f''(x)\) should be positive. This condition ensures that the curve is concave up, resembling a smiley face, confirming a minimum rather than a maximum.
In our exercise, the given point \((6, -5)\) is a local minimum for the quadratic function. By confirming both \(f'(6) = 0\) and evaluating the function value, we assured the conditions for a local minimum were satisfied.
For a function to have a local minimum, the derivative at that point must be zero \(f'(x) = 0\). This zero slope denotes a flat tangent.- In addition to \(f'(x) = 0\), the second derivative \(f''(x)\) should be positive. This condition ensures that the curve is concave up, resembling a smiley face, confirming a minimum rather than a maximum.
In our exercise, the given point \((6, -5)\) is a local minimum for the quadratic function. By confirming both \(f'(6) = 0\) and evaluating the function value, we assured the conditions for a local minimum were satisfied.
Exploring Quadratic Equations
Quadratic equations are polynomials of degree 2 and they form a parabola when graphed. The standard form is \(f(x) = ax^2 + bx + c\), containing various characteristics that define its shape and position.
- The coefficients \(a, b,\) and \(c\) determine the direction and position of the parabola. - The vertex of the parabola (a critical point) can represent either a minimum or maximum based on the coefficient \(a\). If \(a > 0\), the parabola opens upwards (minimum), and if \(a < 0\), it opens downwards (maximum).
For our given function, \(f(x) = x^2 - 12x + 31\), it is in the vertex form \(y = (x-h)^2 + k\), where \(h\) and \(k\) are coordinates of the vertex. We found \(a = -12\) and \(b = 31\) such that the parabola achieves a local minimum at \((6, -5)\), demonstrating a clear and well-defined behavior of quadratic equations.
- The coefficients \(a, b,\) and \(c\) determine the direction and position of the parabola. - The vertex of the parabola (a critical point) can represent either a minimum or maximum based on the coefficient \(a\). If \(a > 0\), the parabola opens upwards (minimum), and if \(a < 0\), it opens downwards (maximum).
For our given function, \(f(x) = x^2 - 12x + 31\), it is in the vertex form \(y = (x-h)^2 + k\), where \(h\) and \(k\) are coordinates of the vertex. We found \(a = -12\) and \(b = 31\) such that the parabola achieves a local minimum at \((6, -5)\), demonstrating a clear and well-defined behavior of quadratic equations.