Problem 9 - Calculus A New Horizon

Question

Find: (a) the intervals on which \(f\) is increasing, (b) the intervals on which \(f\) is decreasing, (c) the open intervals on which \(f\) is concave up. (d) the open intervals on which \(f\) is concave down, and (e) the \(x\) -coordinates of all inflection points. $$f(x)=x^{2}-5 x+6$$

Step-by-Step Solution

Answer

Increasing on \((\frac{5}{2}, fty)\), decreasing on \((- fty, \frac{5}{2})\), concave up on \((-\infty, \infty)\), no intervals are concave down, no inflection points.

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first find the derivative of the function. The given function is \( f(x) = x^2 - 5x + 6 \). Differentiate with respect to \( x \): \[f'(x) = \frac{d}{dx}(x^2 - 5x + 6) = 2x - 5\]

Step 2: Find Critical Points from First Derivative

To find critical points, set the first derivative equal to zero and solve for \( x \):\[2x - 5 = 0\]\[2x = 5\]\[x = \frac{5}{2}\]This is our critical point.

Step 3: Determine Intervals of Increase/Decrease

We determine where \( f'(x) \) is positive or negative by plugging values from intervals around \( x = \frac{5}{2} \) into the first derivative.- Test \( x = 2 \): \( f'(2) = 2(2) - 5 = -1 \), negative implies decreasing.- Test \( x = 3 \): \( f'(3) = 2(3) - 5 = 1 \), positive implies increasing.Thus, \( f(x) \) is decreasing on \((-fty, \frac{5}{2})\) and increasing on \((\frac{5}{2}, fty)\).

Step 4: Find the Second Derivative

To determine the concavity, find the second derivative of the function, \( f(x) = x^2 - 5x + 6 \):\[f''(x) = \frac{d}{dx}(2x - 5) = 2\]

Step 5: Analyze Concavity

Since the second derivative \( f''(x) = 2 \) is positive for all \( x \), this indicates that \( f(x) \) is concave up on \( (-\infty, \infty) \) and there's no interval where it is concave down.

Step 6: Identify Inflection Points

Inflection points occur where the concavity changes, which is where the second derivative changes sign. Since \( f''(x) = 2 \) does not change sign (it's always positive), there are no inflection points.

Key Concepts

Derivatives

The concept of derivatives is foundational in calculus. A derivative represents the rate at which a function changes as its input changes. Essentially, it is the slope of the function at any given point.

For the function given in the exercise, we start by differentiating it to find the first derivative. The function is:

• Given: \( f(x) = x^2 - 5x + 6 \)
• First Derivative: \( f'(x) = 2x - 5 \)

A positive derivative indicates the function is increasing, while a negative one shows it is decreasing. This helps determine how the function behaves over different intervals.

Critical Points

Critical points are crucial for understanding when a function changes its increasing or decreasing behavior. These are points where the first derivative equals zero or does not exist. At these points, the function may have a local maximum, local minimum, or potentially neither.

For our specific exercise, we find a critical point by setting the derivative, \( f'(x) = 2x - 5 \), equal to zero:

• \( 2x - 5 = 0 \)
• \( x = \frac{5}{2} \)

This calculation tells us that at \( x = \frac{5}{2} \), the function may transition from increasing to decreasing, or vice-versa.

Concavity

Concavity tells us how a function curves. If a function is concave up, it's shaped like a smile or a cup. When it's concave down, it's frowning or shaped like a cap. The second derivative provides insight into the concavity of a function.

In this exercise, we've derived the function twice to find the second derivative:

• First, take the derivative of \( f(x) = x^2 - 5x + 6 \) to get \( f'(x) = 2x - 5 \)
• Then, derive \( f'(x) \) to get the second derivative \( f''(x) = 2 \)

Since \( f''(x) = 2 \) is a constant positive value, it implies the function is concave up for all intervals across its domain. There are no intervals of concave down for this function.

Inflection Points

Inflection points occur where there is a change in the direction of concavity. The function changes from being concave up to concave down or vice versa at these points. Inflection points give us a clearer picture of the curve's nature at particular intervals.

For our function in the exercise, we look at the second derivative:

• The second derivative was found to be \( f''(x) = 2 \)
• Since it does not change sign, there is no point where the concavity changes

This shows that the function has no inflection points because it remains concave up throughout its domain. Understanding this helps students see where, if any, twist points in the curve occur.