Question

A function \(f(x)\) is given. Find the \(x\) values where \(f^{\prime}(x)\) has a relative maximum or minimum. \(f(x)=-x^{2}-5 x+7\)

Short answer

The function's first derivative does not have a relative maximum or minimum.

Step-by-step solution

Find the First Derivative

To find where the derivative has a relative maximum or minimum, we first determine the derivative of the function. Given \( f(x) = -x^2 - 5x + 7 \), find the first derivative \( f'(x) \). Differentiate each term to get:\[ f'(x) = \frac{d}{dx}(-x^2) + \frac{d}{dx}(-5x) + \frac{d}{dx}(7) = -2x - 5 \]

Find the Second Derivative

Next, to investigate where \( f'(x) \) has a relative maximum or minimum, find the second derivative of the function using the derivative we obtained.\[ f''(x) = \frac{d}{dx}(-2x - 5) = -2 \]

Determine the Nature of the First Derivative

Observe that the second derivative \( f''(x) = -2 \) is a constant value and never zero. Since it is negative, \( f'(x) = -2x - 5 \) is a decreasing function. This means \( f'(x) \) cannot have a relative maximum or minimum because it does not change sign.

Key concepts

Derivatives

Derivatives are a core concept in calculus used to understand the rate of change of a function. When we talk about differentiating a function, we typically mean finding its derivative, which provides a mathematical way to observe how the function behaves as its input changes. Given a function, say \( f(x) = -x^2 - 5x + 7 \), we compute its first derivative \( f'(x) \) to find the rate at which \( f(x) \) is changing at any point \( x \).

• The power rule helps in differentiating terms like \( -x^2 \) to \( -2x \).
• For linear terms such as \( -5x \), the derivative is simply \( -5 \).
• Constants like \( 7 \) have a derivative of zero because they do not change.

Putting it all together, the derivative for our specific function is \( f'(x) = -2x - 5 \). This expression tells us how the original function \( f(x) \) slopes upwards or downwards across different \( x \) values.

Critical Points

Critical points are where the first derivative of a function is either zero or undefined. These points are important because they are potential locations for relative maxima or minima of a function.To find critical points, set the first derivative \( f'(x) \) equal to zero and solve for \( x \). For our derivative \( f'(x) = -2x - 5 \), it never equals zero due to the constant slope represented by \( -2 \). Therefore, \( -2x - 5 \) does not have critical points.
Understanding critical points gives insight into where functions may peak or dip. However, in this case, given the linear nature of \( f'(x) \), it consistently decreases, leaving no room for maxima or minima. This negative slope indicates that the function itself may have peaks or valleys elsewhere if its derivative changes form.

Second Derivative Test

The second derivative test is a method used to determine whether a critical point is a relative maximum, minimum, or neither. It involves evaluating the second derivative of a function, \( f''(x) \).The logic behind this test is:

• If \( f''(x) > 0 \) at the critical point, the function is concave upward, indicating a relative minimum.
• If \( f''(x) < 0 \) at the critical point, the function is concave downward, indicating a relative maximum.
• If \( f''(x) = 0 \), the test is inconclusive.

In our exercise, the second derivative \( f''(x) = -2 \) is constant and negative, suggesting a concave downward shape at all points. However, since the derivative \( f'(x) \) is a linear function without critical points, the test indicates no relative maxima or minima exist for \( f'(x) \). This reflects the broader nature of the function, advising us that \( f'(x) \) doesn’t have peaks or troughs due to the linear decrease signified by \( f''(x) \).