Question

A function \(f(x)\) is given. Find the \(x\) values where \(f^{\prime}(x)\) has a relative maximum or minimum. . \(f(x)=x^{2}-2 x+1\)

Short answer

The function \( f'(x) \) has a relative minimum at \( x = 1 \).

Step-by-step solution

Find the first derivative

To find the points where the first derivative of the function, \( f(x) \), has a relative maximum or minimum, we first need to find the first derivative. Differentiate \( f(x) = x^2 - 2x + 1 \):\[f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(2x) + \frac{d}{dx}(1) = 2x - 2\]

Determine critical points of the derivative

To find the critical points of the derivative \( f'(x) = 2x - 2 \), we set it equal to zero:\[2x - 2 = 0\]Solving for \( x \):\[2x = 2 \x = 1\]So, \( x = 1 \) is the critical point of the derivative.

Analyze the second derivative

Find the second derivative to analyze the critical points from the first derivative. Differentiate \( f'(x) = 2x - 2 \):\[f''(x) = \frac{d}{dx}(2x - 2) = 2\]

Conclude if the critical point is a max/min

Since the second derivative \( f''(x) = 2 \) is positive, the function \( f(x) = x^2 - 2x + 1 \) is concave up. Thus, the critical point \( x = 1 \) is a minimum point for the derivative \( f'(x) \).

Key concepts

First Derivative

The first derivative of a function, often denoted as \( f'(x) \), provides crucial information about the slope of the tangent to a curve at any point \( x \). Calculating the first derivative is the initial step in identifying potential points of relative maximum or minimum. For our function \( f(x) = x^2 - 2x + 1 \), the first derivative is obtained by differentiating term by term, leading to \( f'(x) = 2x - 2 \). Simplifying derivative expressions helps to easily determine where slopes are zero, indicating possible critical points, which are essential for further analysis.

• Critical points occur where \( f'(x) = 0 \), which implies the slope of the tangent is horizontal, intersecting the x-axis or remaining constant at a specific value of \( x \).
• Critical points can also occur wherever \( f'(x) \) is undefined, but for polynomials like this one, that is not a concern.

To find out more about these critical points, we need to assess both the first and the second derivatives. This leads us into deeper analysis to understand the behavior of the function at these points.

Second Derivative

The second derivative provides insight into the behavior of the graph's concavity. In simple terms, it tells us whether the curve is bending upwards or downwards at a given point. This derivative is especially instrumental in determining the nature of any critical points found using the first derivative. For our function, we already have \( f'(x) = 2x - 2 \). When we differentiate this first derivative again, we find the second derivative: \( f''(x) = 2 \). This value is constant across all \( x \), indicating uniform concavity.

• If \( f''(x) \) is positive, the graph is concave up, resembling a smiley face, implying any critical points are minimums.
• If \( f''(x) \) is negative, the graph is concave down, resembling a frown, suggesting any critical points are maximums.

With our example, since \( f''(x) = 2 \) is positive, the graph of \( f(x) = x^2 - 2x + 1 \) suggests every critical point from the first derivative is a minimum point. Here, this tells us about the nature of the critical point \( x = 1 \), confirming it as a relative minimum for the first derivative \( f'(x) \).

Relative Maximum or Minimum

Determining whether a function has relative maximum or minimum points is a key aspect of calculus, rooted in the behavior and properties of its derivatives. Relative maximum or minimum points describe peaks and valleys on a graph – local high and low points – and these are typically found where the first derivative \( f'(x) \) is zero or undefined.For \( f(x) = x^2 - 2x + 1 \), we calculated \( f'(x) = 2x - 2 \). Setting \( f'(x) = 0 \) gives us \( x = 1 \) as a critical point. This is where we suspect a relative extremum might occur. To confirm, we look at the second derivative.

• A positive second derivative indicates a relative minimum; here, since \( f''(x) = 2 \), which is greater than zero, \( x = 1 \) is a point where the original function has a relative minimum.
• A negative second derivative would denote a relative maximum, but this case doesn't apply here due to the positive second derivative.

So, at \( x = 1 \), the function doesn't just flatten out temporarily – it reaches a valley point relative to the immediate surrounding points. This method and understanding are central parts of determining the curves and shapes of more complex functions.