Problem 12 - Applied Calculus

Question

Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. $$f(x)=x^{2}-5 x+3$$

Step-by-Step Solution

Answer

The critical point is \( x = \frac{5}{2} \), a local minimum; there are no inflection points.

Step 1: Find the First Derivative

To find the critical points, we first need to determine the first derivative of the function. Given \( f(x) = x^2 - 5x + 3 \), the first derivative is \( f'(x) = 2x - 5 \).

Step 2: Solve for Critical Points

Critical points occur where the first derivative is zero or undefined. Set \( f'(x) = 0 \) and solve for \( x \): \[ 2x - 5 = 0 \] \[ 2x = 5 \] \[ x = \frac{5}{2} \].So, the critical point is at \( x = \frac{5}{2} \).

Step 3: Find the Second Derivative

To classify the critical points and find inflection points, we find the second derivative of \( f(x) \). The first derivative is \( f'(x) = 2x - 5 \), so the second derivative is \( f''(x) = 2 \).

Step 4: Classify the Critical Point Using the Second Derivative

Use the second derivative to determine if the critical point is a local maximum, minimum, or neither. Evaluate \( f''(5/2) \):Since \( f''(x) = 2 \) is positive for all \( x \), the function is concave up at \( x = \frac{5}{2} \), indicating a local minimum.

Step 5: Determine Inflection Points

Inflection points occur where the second derivative changes sign. Since \( f''(x) = 2 \) is a constant and does not change sign, there are no inflection points.

Step 6: Verify with a Graph

To confirm our findings, sketch the graph of \( f(x) = x^2 - 5x + 3 \). Observe that the graph is a parabola opening upwards. The vertex, at \( x = \frac{5}{2} \), corresponds to the local minimum found earlier.

Key Concepts

First Derivative

The first derivative of a function is crucial in identifying critical points, a significant part of studying calculus. Here, we had a quadratic function, given by \( f(x) = x^2 - 5x + 3 \). To find the first derivative, simply differentiate the function with respect to \( x \). This gives us \( f'(x) = 2x - 5 \).

This first derivative represents the slope of the tangent line to the curve of the function at any point \( x \). Wherever the first derivative equals zero or is undefined indicates that the slope of the function is either perfectly flat or has some discontinuity.

• This is important because flat slopes often correspond to peaks or valleys – our key critical points.
• The formula \( f'(x) = 0 \) gives us the critical point when it’s solveable; here that happens when \( 2x - 5 = 0 \).

By solving for \( x \), we find that our critical point is at \( x = \frac{5}{2} \). This point is critical because it’s where the function potentially shifts from increasing to decreasing or vice versa.

Critical Points

Once you've identified the first derivative, critical points become the next logical step.

Critical points exist where the derivative is zero or undefined. For our example, with the function \( f(x) = x^2 - 5x + 3 \), we solved \( f'(x)= 0 \), resulting in \( x = \frac{5}{2} \).

In this process, the critical point is where changes in the direction of the function occur.

• Critical points can be classified further into local maxima, local minima, or saddle points.
• To understand what type of point it is, we must evaluate further.

In our case with \( x = \frac{5}{2} \), the critical point requires more to definitively count as a maximum or minimum. For that, a second derivative can confirm.

Second Derivative

The second derivative offers insight into the concavity of the function, which helps determine the nature of critical points. Calculating it involves taking the derivative of the first derivative.

For our function \( f(x) = x^2 - 5x + 3 \), the first derivative was \( f'(x) = 2x - 5 \), which leads to a constant second derivative: \( f''(x) = 2 \).

The constant nature of \( f''(x) = 2 \) has important implications:

• It shows that the function is consistently concave up, as it does not change sign.
• This tells us that any critical point found is at a local minimum, since the second derivative is positive everywhere.

In essence, if \( f''(x) > 0 \), the function has a local minimum at critical points. Conversely, if \( f''(x) < 0 \), you get a local maximum.

Inflection Points

In the study of calculus, inflection points signify where a function changes concavity.

However, if the second derivative remains constant, as it does in \( f''(x) = 2 \), inflection points aren't in the picture.

• Inflection points occur when there is a sign change in the second derivative.
• This means for an inflection point to exist, \( f''(x) \) would need varied positive/negative results over its domain.

Because our example has a constant positive second derivative, \( f(x) = x^2 - 5x + 3 \) stands without inflection points, since the curve's concavity does not change. This is quite common in simple quadratic functions, which generally have the shape of a parabola.