Problem 61 - Calculus Early Transcendentals

Question

a. Find the critical points of \(f\) on the given interval. b. Determine the absolute extreme values of \(f\) on the given interval. c. Use a graphing utility to confirm your conclusions. $$f(x)=x^{3} e^{-x} \text { on }[-1,5]$$

Step-by-Step Solution

Answer

Absolute Minimum: 0 (at x=0) Absolute Maximum: -e (at x=-1)

Step 1: Find the derivative of the function

To find the critical points, we first need to find the derivative of the function, \$f'(x)\$. We can use the product rule and the chain rule to find the derivative of the function. The product rule states that if \$u(x)\$ and \$v(x)\$ are differentiable functions, then the derivative of the product \$u(x)v(x)\$ is given by: $$\frac{d}{dx}[u(x)v(x)]=u'(x)v(x)+u(x)v'(x)$$ Let \$u(x)=x^3\$ and \$v(x)=e^{-x}\$. Then, we can find the derivatives of \$u\$ and \$v\$: $$u'(x)=3x^2$$ $$v'(x)=-e^{-x}$$ Now, use the product rule to find the derivative of \$f(x)\$: $$f'(x)=u'(x)v(x)+u(x)v'(x)=3x^2e^{-x}+x^3(-e^{-x})$$

Step 2: Find the critical points

To find the critical points of \$f(x)\$, we need to set the derivative equal to zero and solve for \$x\$: $$f'(x)=0=3x^2e^{-x}+x^3(-e^{-x})$$ Factor out an \$x^2e^{-x}\$ term: $$0=x^2e^{-x}(3-x)$$ From this equation, we can see that there are two critical points: \$x=0\$ and \$x=3\$. However, we need to check if these critical points are within the given interval \[-1,5\]. Both \$0\$ and \$3\$ are within the given interval, so these are the critical points we'll consider when finding the absolute extreme values in the next step.

Step 3: Determine the absolute extreme values

To find the absolute extreme values of the function, we need to evaluate \$f(x)\$ at the critical points and the endpoints of the interval. That means we'll evaluate \$f(x)\$ at \$x=-1\$, \$x=0\$, \$x=3\$, and \$x=5\$. Then, we'll compare their values. 1. \$f(-1)=(-1)^3e^1=-e\$ 2. \$f(0)=(0)^3e^0=0\$ 3. \$f(3)=(3)^3e^{-3}=27e^{-3}\$ 4. \$f(5)=(5)^3e^{-5}=125e^{-5}\$ The absolute minimum value occurs at \$x=0\$ with a value of 0, and the absolute maximum value occurs at \$x=-1\$ with a value of \$-e\$.

Step 4: Use a graphing utility to confirm conclusions

Use a graphing utility, such as Desmos or a graphing calculator, to plot the function \$f(x)=x^3e^{-x}\$ on the interval \[-1,5\]. Observe that the graph confirms the conclusions from Step 3 - the absolute minimum occurs at \$x=0\$ and the absolute maximum occurs at \$x=-1\$.

Key Concepts

Product Rule

The Product Rule is a vital tool in calculus, allowing us to differentiate products of functions. If we have two functions, say \( u(x) \) and \( v(x) \), the rule states that the derivative of their product \( u(x)v(x) \) is:

• \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

For the function \( f(x) = x^3 e^{-x} \), we identify:

• \( u(x) = x^3 \times v(x) = e^{-x} \)

Thus,:\( u'(x) = 3x^2 \) and \( v'(x) = -e^{-x} \). Applying the product rule gives us the derivative \( f'(x) = 3x^2 e^{-x} + x^3 (-e^{-x}) \). This allows us to find critical points by setting \( f'(x) = 0 \) and solving for \( x \).Understanding how the product rule breaks down a complex function into manageable parts makes calculus much easier.

Chain Rule

The Chain Rule helps us differentiate composite functions, where one function is nested inside another. It's expressed as:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) \)

In our example, even though the primary focus is on the Product Rule, recognizing how the Chain Rule would apply is useful. Consider \( v(x) = e^{-x} \), where \( e^u \) is the outer function and \( u = -x \) is the inner function. Thus, the derivative \( v'(x) = e^{-x}(-1) = -e^{-x} \). The Chain Rule quickly gives derivatives of nested functions, essential for solving complex problems.

Absolute Extreme Values

To find the absolute extreme values of a function on an interval, we evaluate the function at its critical points and endpoints. These values tell us where the function reaches its highest and lowest points. In this example, we evaluated \( f(x) = x^3 e^{-x} \) at critical points \( x = 0 \) and \( x = 3 \), and endpoints \( x = -1 \) and \( x = 5 \):

• \( f(-1) = -e \)
• \( f(0) = 0 \)
• \( f(3) = 27e^{-3} \)
• \( f(5) = 125e^{-5} \)

The absolute minimum is at \( x = 0 \) with value 0, and the maximum is at \( x = -1 \) with value \( -e \). This systematic evaluation gives us a clear picture of the function’s behavior.

Interval Notation

Interval notation provides a compact way to express a range of values. It's specially helpful in calculus when defining the domain of interest. In this exercise, the interval is \([-1, 5]\), which means:

• The interval includes \(-1\) and \(5\), indicated by the brackets \([ \text{} ]\).

If it were open, say \((a, b)\), the parentheses \(( \text{} )\) would indicate exclusion of the endpoints. Using this notation helps clearly define where to check for critical points and to evaluate extremes. It gives a precise boundary for functions, which is crucial for proper analysis of their behavior across specific domains.