Problem 22 - Essential Calculus

Question

Find the critical numbers of \(f\) (if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results. $$ f(x)=x^{3}-6 x^{2}+15 $$

Step-by-Step Solution

Answer

The critical numbers of the function are x=0 and x=4. The function is increasing for \(x<0\) and \(x>4\), and decreasing for \(0

Step 1: Derive the function

First, you need to find the derivative of the function \(f(x)=x^{3}-6 x^{2}+15\). The derivative of \(x^{3}\) is \(3x^{2}\), of \(-6x^{2}\) is \(-12x\), and the derivative of a constant (15) is 0. So, the derivative of the function \(f'(x)= 3x^{2}-12x\).

Step 2: Solve the derivative equals to zero

Next, you need to set this derivative equal to zero and solve for \(x\), to find critical numbers. So, you have the equation \(3x^{2}-12x=0\). By factoring out \(3x\), you get \(3x(x-4)=0\). So, the solutions to this equation (the critical numbers) are \(x=0\) and \(x=4\).

Step 3: Test intervals

Then test the sign of the derivative in each of the intervals determined by the critical numbers: (-Infinity,0), (0,4), and (4, Infinity). Take any number within these intervals and put into the derivative.Take -1 for (-Infinity,0), we have \(3(-1)^2-12(-1) >0 \), so \(f(x)\) is increasing in this interval.For interval (0,4), take x=2 then we have \(3(2)^2-12(2) <0 \), hence \(f(x)\) is decreasing in this interval.For interval (4,Infinity), take x=5 then we have \(3(5)^2-12(5) >0 \), so \(f(x)\) is increasing.

Step 4: Identify relative extrema

Given that the function changes from increasing to decreasing at x=0, this point is a relative maximum. Conversely, since the function changes from decreasing to increasing at x=4, there is a relative minimum at this point.

Step 5: Confirm with a graph

Plot the function \(f(x) = x^{3} - 6x^{2} + 15\) to observe the maximum at x=0 and the minimum at x=4, and that the function is increasing for \(x<0\) and \(x>4\) and decreasing for \(0

Key Concepts

Derivative of Polynomial

When we are working with polynomial functions like \( f(x) = x^{3} - 6x^{2} + 15 \), understanding derivatives is essential. The derivative represents the rate at which the function's value is changing at any given point on its graph. To calculate the derivative of a polynomial, we apply the power rule, which involves multiplying the coefficient of each term by its exponent and then decreasing the exponent by one.

For our given function \( f(x) \), the power rule gives us the derivative \( f'(x) = 3x^{2} - 12x \). This new function tells us how the slope of the original function changes as \( x \) varies. Moreover, it is critical in finding critical numbers, which are the values of \( x \) where \( f'(x) = 0 \) or where the derivative does not exist. In this particular polynomial, there are no points where the derivative does not exist, so we simply solve \( f'(x) = 0 \) to find the critical numbers.

Relative Extrema

Relative extrema are the peaks and valleys — the high and low points — on the graph of a function. We can determine these by analyzing the critical numbers found by setting the derivative equal to zero. To find extrema, we need to check where the function changes from increasing to decreasing, or vice versa, at these critical numbers.

In the solution for our function \( f(x) \), we observe that the function goes from increasing to decreasing at \( x=0 \), and from decreasing to increasing at \( x=4 \). This means that at \( x=0 \) we have a relative maximum since the value of the function is higher than it is immediately before or after that point. Conversely, at \( x=4 \), we have a relative minimum because the value of the function is less than around that point. Confirming these findings with a graphing utility can help us visualize the behavior of the function and solidify our understanding of relative extrema.

Intervals of Increase and Decrease

The intervals of increase and decrease tell us where the function is moving upwards or downwards as we read the graph from left to right. After finding the critical numbers and determining the sign of the derivative in the regions divided by these numbers, we can identify these intervals. For the function \( f(x) = x^{3} - 6x^{2} + 15 \), we test values within each interval to see if the function is increasing or decreasing.

Through testing, we find that when \( x < 0 \) and \( x > 4 \), the derivative is positive, which means that \( f(x) \) is increasing. Conversely, for the interval from \( 0 < x < 4 \), the derivative is negative, indicating that \( f(x) \) is decreasing. These intervals have deep implications in visualizing the shape of the graph and understanding the behavior of the function across its domain.

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Other exercises in this chapter

Problem 22

In Exercises \(15-36,\) find the limit. $$ \lim _{x \rightarrow-\infty} \frac{x}{\sqrt{x^{2}+1}} $$

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Problem 22

Use a graphing utility to graph the function on the closed interval [a,b]. Determine whether Rolle's Theorem can be applied to \(f\) on the interval and, if so,

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Problem 22

Locate the absolute extrema of the function on the closed interval. $$ y=3-|t-3|,[-1,5] $$

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Problem 23

Writing In Exercises 23 and 24 , give a short explanation of why the approximation is valid. $$ \sqrt{4.02} \approx 2+\frac{1}{4}(0.02) $$

View solution

Key Concepts

Derivative of Polynomial

When we are working with polynomial functions like \( f(x) = x^{3} - 6x^{2} + 15 \), understanding derivatives is essential. The derivative represents the rate at which the function's value is changing at any given point on its graph. To calculate the derivative of a polynomial, we apply the power rule, which involves multiplying the coefficient of each term by its exponent and then decreasing the exponent by one.

For our given function \( f(x) \), the power rule gives us the derivative \( f'(x) = 3x^{2} - 12x \). This new function tells us how the slope of the original function changes as \( x \) varies. Moreover, it is critical in finding critical numbers, which are the values of \( x \) where \( f'(x) = 0 \) or where the derivative does not exist. In this particular polynomial, there are no points where the derivative does not exist, so we simply solve \( f'(x) = 0 \) to find the critical numbers.

Relative Extrema

Relative extrema are the peaks and valleys — the high and low points — on the graph of a function. We can determine these by analyzing the critical numbers found by setting the derivative equal to zero. To find extrema, we need to check where the function changes from increasing to decreasing, or vice versa, at these critical numbers.

In the solution for our function \( f(x) \), we observe that the function goes from increasing to decreasing at \( x=0 \), and from decreasing to increasing at \( x=4 \). This means that at \( x=0 \) we have a relative maximum since the value of the function is higher than it is immediately before or after that point. Conversely, at \( x=4 \), we have a relative minimum because the value of the function is less than around that point. Confirming these findings with a graphing utility can help us visualize the behavior of the function and solidify our understanding of relative extrema.

Intervals of Increase and Decrease

The intervals of increase and decrease tell us where the function is moving upwards or downwards as we read the graph from left to right. After finding the critical numbers and determining the sign of the derivative in the regions divided by these numbers, we can identify these intervals. For the function \( f(x) = x^{3} - 6x^{2} + 15 \), we test values within each interval to see if the function is increasing or decreasing.

Through testing, we find that when \( x < 0 \) and \( x > 4 \), the derivative is positive, which means that \( f(x) \) is increasing. Conversely, for the interval from \( 0 < x < 4 \), the derivative is negative, indicating that \( f(x) \) is decreasing. These intervals have deep implications in visualizing the shape of the graph and understanding the behavior of the function across its domain.