Problem 15
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. $$f(x)=x^{2}-3 x+8$$
Step-by-Step Solution
Verified Answer
(a) Increasing on \((\frac{3}{2}, \infty)\), (b) Decreasing on \((-\infty, \frac{3}{2})\), (c) Concave up on \((-\infty, \infty)\), (d) No concave down intervals, (e) No inflection points.
1Step 1: Find the First Derivative
To determine where the function \( f(x) = x^2 - 3x + 8 \) is increasing or decreasing, we first find its first derivative \( f'(x) \). Using the power rule, we differentiate each term: \( f'(x) = 2x - 3 \).
2Step 2: Determine Critical Points
To find the critical points, set the first derivative equal to zero: \( 2x - 3 = 0 \). Solving for \( x \), we get: \( x = \frac{3}{2} \), which is our critical point.
3Step 3: Test Intervals for Increasing/Decreasing Function
We create test intervals using the critical point \( x = \frac{3}{2} \):- Interval 1: \((-\infty, \frac{3}{2})\)- Interval 2: \((\frac{3}{2}, \infty)\)Choose test points from each interval, such as \( x = 0 \) for Interval 1 and \( x = 2 \) for Interval 2.Calculate \( f'(0) = 2(0) - 3 = -3 \). Since \( f'(0) < 0 \), the function is decreasing on \((-\infty, \frac{3}{2})\).Calculate \( f'(2) = 2(2) - 3 = 1 \). Since \( f'(2) > 0 \), the function is increasing on \((\frac{3}{2}, \infty)\).
4Step 4: Find the Second Derivative
To determine concavity, we find the second derivative of the function. Differentiate \( f'(x) = 2x - 3 \) to get \( f''(x) = 2 \).
5Step 5: Analyze Concavity
Since \( f''(x) = 2 \) is a constant positive number, the function is concave up on all intervals. Therefore, it is concave up on \((-\infty, \infty)\) and concave down on none.
6Step 6: Identify Inflection Points
Inflection points occur where the second derivative changes sign. Since \( f''(x) = 2 \) does not change sign and is always positive, there are no inflection points.
Key Concepts
First DerivativeCritical PointsConcavityInflection Points
First Derivative
The first derivative is a powerful tool in calculus used to determine how a function behaves. The first derivative, denoted as \( f'(x) \), tells us the rate at which the function \( f(x) \) changes at any given point. In simpler terms, it's like the speed of the function: how fast or slow it grows or shrinks.
For the function \( f(x) = x^2 - 3x + 8 \), calculating the first derivative involves applying the power rule. This rule allows us to differentiate each term separately. Here, \( f'(x) = 2x - 3 \). What this tells us is how the slope of the function behaves at different points on the graph.
To find where the function is increasing or decreasing, you set \( f'(x) = 0 \) to find the critical points. These points are where the function could potentially have a maximum, minimum, or neither. The critical point, found at \( x = \frac{3}{2} \), splits the graph into regions where we can analyze the function's behavior using a test point from each interval.
For the function \( f(x) = x^2 - 3x + 8 \), calculating the first derivative involves applying the power rule. This rule allows us to differentiate each term separately. Here, \( f'(x) = 2x - 3 \). What this tells us is how the slope of the function behaves at different points on the graph.
To find where the function is increasing or decreasing, you set \( f'(x) = 0 \) to find the critical points. These points are where the function could potentially have a maximum, minimum, or neither. The critical point, found at \( x = \frac{3}{2} \), splits the graph into regions where we can analyze the function's behavior using a test point from each interval.
Critical Points
Critical points are parts of a function's graph where the derivative is zero or undefined. They are essential for identifying the "hills" or "valleys" of a function, where it changes direction or flattens out.
For our function \( f(x) = x^2 - 3x + 8 \), the first derivative \( f'(x) = 2x - 3 \) equals zero at \( x = \frac{3}{2} \). This critical point divides our function into sections for further evaluation of increase or decrease.
To decide if the function is increasing or decreasing at these sections, you'll evaluate \( f'(x) \) using test points from intervals on either side of this critical point:
For our function \( f(x) = x^2 - 3x + 8 \), the first derivative \( f'(x) = 2x - 3 \) equals zero at \( x = \frac{3}{2} \). This critical point divides our function into sections for further evaluation of increase or decrease.
To decide if the function is increasing or decreasing at these sections, you'll evaluate \( f'(x) \) using test points from intervals on either side of this critical point:
- From \(( -\infty, \frac{3}{2})\), pick a number like \( x = 0 \) and evaluate. The result is negative, indicating the function decreases.
- From \(( \frac{3}{2}, \infty )\), try \( x = 2 \). The result is positive, so the function increases.
Concavity
Concavity in calculus describes how a function curves. To find it, the second derivative, \( f''(x) \), is used. While the first derivative tells us about slope, the second derivative informs us about the curve's direction.
Calculating the second derivative of our function \( f'(x) = 2x - 3 \), we get \( f''(x) = 2 \). This is a constant value, meaning there are no changes in concavity.
Since \( f''(x) \) is positive in all sections (always 2), our function is concave up on the interval \(( -\infty, \infty )\), unmistakably "cup-shaped" at every point. When concavity doesn’t switch between positive and negative, as in our function, it doesn't have sections that are "concave down."
Thus, our function bends upwards consistently across its entire domain.
Calculating the second derivative of our function \( f'(x) = 2x - 3 \), we get \( f''(x) = 2 \). This is a constant value, meaning there are no changes in concavity.
Since \( f''(x) \) is positive in all sections (always 2), our function is concave up on the interval \(( -\infty, \infty )\), unmistakably "cup-shaped" at every point. When concavity doesn’t switch between positive and negative, as in our function, it doesn't have sections that are "concave down."
Thus, our function bends upwards consistently across its entire domain.
Inflection Points
An inflection point is where the graph changes its concavity. For an inflection point to exist, the second derivative \( f''(x) \) must change signs. It marks where the graph shifts from curving "up" to "down" or vice versa.
For the given function \( f(x) = x^2 - 3x + 8 \), its second derivative \( f''(x) = 2 \) stayed constant and positive. This indicates no inflection point because the concavity doesn't change across the entire domain.
Without a change in the sign of \( f''(x) \), there isn't a place where the graph shifts its shape from concave up to concave down, or vice versa. Hence, our function lacks inflection points as it maintains that upward curve consistently.
For the given function \( f(x) = x^2 - 3x + 8 \), its second derivative \( f''(x) = 2 \) stayed constant and positive. This indicates no inflection point because the concavity doesn't change across the entire domain.
Without a change in the sign of \( f''(x) \), there isn't a place where the graph shifts its shape from concave up to concave down, or vice versa. Hence, our function lacks inflection points as it maintains that upward curve consistently.
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