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Chapter 4: Problem 14

Find the exact global maximum and minimum values of the function. The domain is all real numbers unless otherwise specified. $$g(x)=4 x-x^{2}-5$$

Short Answer

Expert verified
Global maximum is -1 at x=2; no finite global minimum (heads to -鈭�).

Step by step solution

01

Find the Derivative

First, find the derivative of the function to determine where the critical points are located. The function is given as \(g(x) = 4x - x^2 - 5\). The derivative, \(g'(x)\), is calculated as follows: \[ g'(x) = \frac{d}{dx}(4x - x^2 - 5) = 4 - 2x \]
02

Solve for Critical Points

Set the derivative equal to zero to find critical points. \[ 4 - 2x = 0 \] Solving for \(x\) gives: \[ 2x = 4 \] \[ x = 2 \]
03

Evaluate the Function at Critical Points

Substitute the critical point back into the original function to find the corresponding \(y\) value.\[ g(2) = 4(2) - (2)^2 - 5 = 8 - 4 - 5 = -1 \]
04

Analyze the Second Derivative

Find the second derivative to determine the concavity at the critical point.First, compute the second derivative:\[ g''(x) = \frac{d}{dx}(4 - 2x) = -2 \]Since \(g''(x) = -2 < 0\), the function is concave down, indicating a local (and, for a quadratic function, also global) maximum at \(x = 2\).
05

Determine Global Behavior

Since \(g(x) = 4x - x^2 - 5\) is a downward-facing parabola (as indicated by the negative coefficient of \(x^2\)), the point \((2, -1)\) is the highest point on the entire curve, which makes it the global maximum.As \(x\) approaches positive or negative infinity, the \(x^2\) term dominates, and the function heads towards negative infinity, indicating no global minimum within the real number domain but the function coinciding with negative infinity as \(x\) approaches \(\pm\infty\).

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Key Concepts

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

Understanding Derivatives
Derivatives provide a way to understand how a function changes at any given point. When you have a function, the derivative tells you the rate at which the function's value is changing with respect to its input, often denoted as \(x\).
For example, from the function \(g(x) = 4x - x^2 - 5\), calculating the derivative involves determining the slope of this function at any point \(x\).
  • The derivative of a constant, like \(-5\), is 0 because constants don't change as \(x\) changes.
  • For \(4x\), the derivative becomes 4, since it's a linear term.
  • The derivative of \(-x^2\) is \(-2x\), reflecting how the squared term changes with \(x\).
Thus, the derivative here is \(g'(x) = 4 - 2x\). By understanding derivatives, you can quickly know how your function behaves across the domain.
Exploring Quadratic Functions
Quadratic functions are polynomial functions of degree two, and they have the general form \(ax^2 + bx + c\). The function \(g(x) = 4x - x^2 - 5\) can be classified as a quadratic function.
This specific function is slightly out of the usual order, but it follows the properties of a quadratic function:
  • It has an \(x^2\) term, meaning it's parabolic in shape.
  • The leading coefficient (\(-1\) for \(-x^2\)) is negative, indicating the parabola opens downward.
  • Quadratic functions are symmetric about their vertex, which is either a maximum or a minimum point.
Understanding that this is a quadratic function helps predict behavior, like determining that any appearance of a maximum means it is the global maximum when the parabola opens downward.
Identifying Critical Points
Critical points of a function occur where the derivative is zero or undefined. These points indicate where the function might change direction.
For the function \(g(x) = 4x - x^2 - 5\) with its derivative \(g'(x) = 4 - 2x\), we find critical points by solving \(g'(x) = 0\). This method helps identify potential maxima, minima, or saddle points.
  • Set \(4 - 2x = 0\).
  • Solve to get \(x = 2\) as the critical point.
With this knowledge, we further analyze the function's behavior around the critical point by evaluating the function value at \(x = 2\), confirming the existence of either a maximum or a minimum at that point.
Significance of the Second Derivative
The second derivative of a function gives insights into its concavity, which helps determine whether a critical point is a maximum or a minimum. If \(g''(x) > 0\), the function is concave up, suggesting a minimum. Conversely, if \(g''(x) < 0\), the function is concave down, signaling a maximum.
In our function \(g(x) = 4x - x^2 - 5\), the second derivative is calculated as \(g''(x) = -2\).
  • Since \(g''(x) = -2 < 0\), it confirms that the function is concave down.
  • This confirms the point \((2, -1)\) is a local maximum.
For a quadratic function like this one, the local maximum is also the global maximum, simplifying our search for extreme values of the function across its entire domain.

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