Problem 73 - University Calculus: Early Transcendentals

Question

Determine the values of constants \(a\) and \(b\) so that \(f(x)=\) \(a x^{2}+b x\) has an absolute maximum at the point (1,2)

Step-by-Step Solution

Answer

The constants are \( a = -2 \) and \( b = 4 \).

Step 1: Understand the problem

We are given a quadratic function \( f(x) = ax^2 + bx \) and we need to find the constants \( a \) and \( b \) such that there is an absolute maximum at the point (1,2). The point (1,2) tells us that when \( x = 1 \), \( f(x) = 2 \) and it's the highest value, hence \( f'(x) = 0 \) at \( x = 1 \).

Step 2: Set up equations for the vertex condition

The vertex of a quadratic in the form \( ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). Given that the maximum occurs at \( x = 1 \), we have:\[-\frac{b}{2a} = 1.\]

Step 3: Solve vertex condition equation

From the equation \(-\frac{b}{2a} = 1\), we solve for \( b \) in terms of \( a \):\[b = -2a.\]

Step 4: Apply the function value condition

At the point (1, 2), the function should equal 2:\[ f(1) = a(1)^2 + b(1) = 2.\]Substitute \( b = -2a \) into the equation:\[ a - 2a = 2. \]

Step 5: Solve for a

Solve \( a - 2a = 2 \):\[-a = 2,\]\[a = -2.\]

Step 6: Solve for b

Use \( b = -2a \) to find \( b \):\[ b = -2(-2) = 4. \]

Step 7: Verify the solution

Substitute \( a = -2 \) and \( b = 4 \) into \( f(x) = ax^2 + bx \) and ensure the function has a maximum at \( (1,2) \).\[ f(x) = -2x^2 + 4x,\]\[ f'(x) = -4x + 4. \]The critical point, \( f'(x) = 0 \), gives \( x = 1 \) as expected and substitution shows \( f(1)=2 \). The function is a downward-facing parabola so the point is indeed a maximum.

Key Concepts

Understanding Absolute Maximum in Quadratic Functions

In the context of a quadratic function, an absolute maximum is the highest point on the graph of the function. For quadratic functions like \( f(x) = ax^2 + bx \), this occurs when the parabola opens downward, resembling an upside-down "U" shape. The absolute maximum is found at the vertex of the parabola, where the slope changes from positive to negative.
The key condition to identify an absolute maximum at a specific point \((x, f(x))\) is that the derivative, \( f'(x) \), at this point must equal zero. This signifies a horizontal tangent or turning point. If the value at the vertex is the highest attainable value, that's an absolute maximum.
In our exercise, to achieve the condition of an absolute maximum at given point \((1,2)\), the function \( f(x) = ax^2 + bx \) must satisfy \( f'(1) = 0 \) and \( f(1) = 2 \). This ensures that the vertex at \( x = 1 \) is the highest value of the function, creating the absolute maximum.

Vertex of a Parabola

The vertex of a parabola in the equation \( ax^2 + bx + c \) is the point of symmetry and can be found using the formula \( x = -\frac{b}{2a} \). This formula helps locate the x-coordinate of the vertex, which is crucial for both maximum and minimum values in quadratic functions.

• The vertex represents the turning point of the parabola.
• It is crucial in determining the nature of the maximum or minimum point depending on the orientation of the parabola.

When focused on maximizing or minimizing a quadratic function, understanding how to derive and use the vertex form can solve many related problems efficiently. For this exercise, setting the vertex \( x = -\frac{b}{2a} = 1 \) implies a specific condition for \( a \) and \( b \), simplifying the determination of these constants.

Function Differentiation and its Role in Quadratic Functions

Differentiation involves finding the derivative of a function, which reflects the rate at which function values change as their input changes. For a quadratic function like \( f(x) = ax^2 + bx + c \), differentiation can pinpoint critical points where the function may achieve a maximum or minimum value.
The first step, finding the first derivative, \( f'(x) \), involves applying basic differentiation rules:

• The derivative of \( ax^2 \) is \( 2ax \).
• The derivative of \( bx \) is just \( b \).

Therefore, \( f'(x) = 2ax + b \).
Setting \( f'(x) = 0 \) reveals the critical point where the function stops increasing and starts decreasing, or vice versa. In this exercise, when \( f'(x) = 0 \) at \( x=1 \), it confirmed that this point might be either a maximum or minimum. By setting these conditions in our problem, we calculate the constants \( a \) and \( b \) that position the maximum at the desired point. Simplified, differentiation converts complex function conditions into manageable equations to solve.