Let's solve each part step by step. The main functions given are: - \( f(x) = 3 \sin x \cos x^2 \) - \( F(x) = \int_0^x f(t) \, dt \) - \( w(x) = \int_0^x F(t) \, dt \) --- ### (a) **Explain why the function \( F(x) \) is nonnegative.** **Solution:** \( F(x) = \int_0^x f(t) \, dt \) is not guaranteed to be nonnegative for all \( x \) unless \( f(t) \geq 0 \) for all \( t \) in the interval \( [0, x] \). But \( f(x) = 3 \sin x \cos x^2 \) can take negative values because both \( \sin x \) and \( \cos x^2 \) oscillate between positive and negative values. Thus, there can be regions where \( f(x) < 0 \), and as a result, \( F(x) \) can also be negative for some \( x \). **So, \( F(x) \) is not necessarily nonnegative.** --- ### (b) **Identify the points of the graph of \( F(x) \) that correspond to the extrema of \( f(x) \). Why is this so? Justify analytically.** **Solution:** *Recall:* Extrema of \( f(x) \) occur where \( f'(x) = 0 \). But **the critical points (max/min) of \( F(x) \) are where its derivative is zero**, i.e., \[ F'(x) = f(x) = 0 \] So **the zeros of \( f(x) \) correspond to the extrema (maxima/minima) of \( F(x) \)**. **Reason:** By the Fundamental Theorem of Calculus, \( F'(x) = f(x) \). So \( F(x) \) is increasing where \( f(x) > 0 \), decreasing where \( f(x) < 0 \), and has a local extremum where \( f(x) = 0 \). --- ### (c) **Does every zero of \( f(x) \) correspond to an extremum of \( F(x) \)? Explain.** **Solution:** A zero of \( f(x) \) is a candidate for an extremum of \( F(x) \), but **not every zero is necessarily a local maximum or minimum**. Some zeros could be inflection points (where the sign of \( f(x) \) does not change as \( x \) passes through the zero). - If \( f(x) \) changes sign at the zero, \( F(x) \) will have a local extremum there. - If \( f(x) \) just touches the axis (i.e., has a double root), then \( F(x) \) may have an inflection point. --- ### (d) **Plot the function \( w(x) = \int_0^x F(t) \, dt \) in the same window as the two functions above. What is the relationship between \( w(x) \) and \( F(x) \)? Justify analytically.** **Solution:** By the Fundamental Theorem of Calculus: - \( w'(x) = F(x) \) - \( F'(x) = f(x) \) So, - \( w(x) \) is an antiderivative of \( F(x) \). - \( F(x) \) is an antiderivative of \( f(x) \). **Relationship:** \( F(x) \) is the first derivative of \( w(x) \): \( w'(x) = F(x) \). \( f(x) \) is the second derivative of \( w(x) \): \( w''(x) = f(x) \). **Summary Table:** | Function | Derivative | |----------|--------------------------| | \( w(x) \) | \( w'(x) = F(x) \) | | \( F(x) \) | \( F'(x) = f(x) \) | --- If you need the explicit plot, you can use Desmos or any graphing calculator to plot these three functions for a specific interval. --- ## **Final Answers (Summary):** **a.** \( F(x) \) is not necessarily nonnegative because \( f(x) \) can be negative. **b.** The extrema of \( F(x) \) occur at the zeros of \( f(x) \), because \( F'(x) = f(x) \). **c.** Not every zero of \( f(x) \) is an extremum of \( F(x) \); only those where \( f(x) \) changes sign. **d.** \( w(x) = \int_0^x F(t) dt \) is an antiderivative of \( F(x) \), so \( w'(x) = F(x) \), and \( w''(x) = f(x) \). The relationship is that \( F(x) \) is the derivative of \( w(x) \).

Let's analyze the problem carefully, step by step, with detailed reasoning and calculations. --- ### Given functions: - \( f(x) = 3 \sin x \cos x^2 \) - \( F(x) = \int_0^x f(t) dt \) - \( w(x) = \int_0^x F(t) dt \) --- ### Part (a): **Explain why the function \( F(x) \) is nonnegative.** **Step 1:** Recall the definition: \[ F(x) = \int_0^x f(t) dt \] **Step 2:** Analyze the sign of \(f(t)\). - \(f(t) = 3 \sin t \cos t^2\). - Both \(\sin t\) and \(\cos t^2\) oscillate between \(-1\) and \(1\). - Therefore, \(f(t)\) can be positive or negative depending on \(t\). **Conclusion:** - Because \(f(t)\) can be negative, the integral \(F(x)\) can increase or decrease as \(x\) varies. - **Hence, \(F(x)\) is *not necessarily* non-negative.** **Final statement:** > **\(F(x)\) is *not guaranteed* to be non-negative because the integrand \(f(t)\) can take negative values.** --- ### Part (b): **Identify the points of the graph of \( F(x) \) that correspond to the extrema of \( f(x) \). Why is this so? Justify analytically.** **Step 1:** Recall the Fundamental Theorem of Calculus: \[ F'(x) = f(x) \] **Step 2:** The extrema (maxima or minima) of \(F(x)\) occur where: \[ F'(x) = 0 \quad \Rightarrow \quad f(x) = 0 \] **Step 3:** These are the points where \(f(x)\) crosses zero, and the slope of \(F\) changes sign. **Step 4:** *Why do these points correspond to extrema?* - At points where \(f(x) = 0\): - \(F'(x) = 0\), so \(F(x)\) has a critical point there. - To determine whether it's a maximum or minimum, analyze the sign of \(f(x)\): - If \(f(x)\) changes from positive to negative, \(F(x)\) has a local maximum at that point. - If \(f(x)\) changes from negative to positive, \(F(x)\) has a local minimum. **Summary:** > **The points where \(f(x) = 0\) correspond to extrema of \(F(x)\) because at these points, the derivative \(F'(x)\) changes sign, indicating a local maximum or minimum.** --- ### Part (c): **Does every zero of \(f(x)\) correspond to an extremum of \(F(x)\)? Explain.** **Step 1:** Consider a zero of \(f(x)\). - At \(x = x_0\), \(f(x_0) = 0\). **Step 2:** For an extremum of \(F(x)\), the necessary condition is \(F'(x_0) = 0\). - This is satisfied here. **Step 3:** However, whether \(F(x)\) has a *local extremum* at \(x_0\) depends on the sign change of \(f(x)\): - If \(f(x)\) changes sign at \(x_0\): - \(F(x)\) has a local maximum or minimum at \(x_0\). - If \(f(x)\) does *not* change sign (touches zero but remains positive or negative): - Then \(F(x)\) has a flat point, but not necessarily an extremum (it could be an inflection point). **Conclusion:** > **Not every zero of \(f(x)\) guarantees an extremum of \(F(x)\). Zeroes where \(f(x)\) does not change sign are points of inflection or flat points, not extrema.** --- ### Part (d): **Plot the function \( w(x) = \int_0^x F(t) dt \). What is its relationship to \(F(x)\)?** **Step 1:** Recall from calculus: \[ w'(x) = F(x) \] and \[ F'(x) = f(x) \] **Step 2:** Therefore: - \(w(x)\) is an *antiderivative* of \(F(x)\). - \(F(x)\) is the *derivative* of \(w(x)\). **Step 3:** **Relationship:** \[ \boxed{ \text{Since } w'(x) = F(x), } \] and \[ \text{since } F'(x) = f(x), \] then \[ w''(x) = F'(x) = f(x). \] **Summary:** - \(w(x)\) is a *second integral* (double integral) of \(f(x)\). - The functions are related via derivatives: \[ \frac{d}{dx} w(x) = F(x), \quad \frac{d}{dx} F(x) = f(x). \] --- ### **Final summarized answers:** | Part | Explanation | Final Statement | |---------|--------------|-------------------| | (a) | Since \(f(t)\) oscillates and can be negative, the integral \(F(x)\) can decrease, becoming negative. | **\(F(x)\) is *not necessarily* nonnegative.** | | (b) | \(F'(x) = f(x)\). Extrema of \(F(x)\) occur where \(f(x) = 0\). | **Points where \(f(x) = 0\) correspond to extrema of \(F(x)\).** | | (c) | Zeroes of \(f(x)\) may not always be extrema unless \(f(x)\) changes sign there. | **Only zeros where \(f(x)\) changes sign correspond to extrema; otherwise, points of inflection.** | | (d) | \(w'(x) = F(x)\), and \(w''(x) = f(x)\). | **\(w(x)\) is an antiderivative of \(F(x)\); the functions are related via derivatives: \(w' = F\) and \(F' = f\).** | --- **Note:** To visualize these functions, you can plot \(f(x)\), \(F(x)\), and \(w(x)\) over the relevant interval, noting how zeros and sign changes relate to extrema and inflection points.

Certainly! While I cannot generate images directly here, I can guide you on how to plot these functions using graphing tools like Desmos, GeoGebra, or a graphing calculator. ### Step-by-step instructions to plot the functions: --- ### 1. Plot \(f(x) = 3 \sin x \cos x^2\) - Enter the function as is. - Observe where \(f(x)\) crosses zero. - Note the oscillatory nature due to the sine and cosine functions. --- ### 2. Plot \(F(x) = \int_0^x f(t) dt\) - In Desmos, you can input: ``` F(x) = \int_0^x 3 \sin t \cos t^2 dt ``` - This will plot the cumulative area under \(f(t)\) from 0 to \(x\). - Alternatively, approximate numerically or use an integral calculator if your tool doesn't support direct integral input. --- ### 3. Plot \(w(x) = \int_0^x F(t) dt\) - Similarly, input: ``` w(x) = \int_0^x F(t) dt ``` - This shows the second antiderivative, which is related to the integral of \(F(t)\). --- ### 4. Analyze the graphs: - **Zeros of \(f(x)\):** where \(f(x)\) crosses the x-axis. These points indicate potential extrema for \(F(x)\). - **Extrema of \(F(x)\):** occur at these zeros where \(f(x)\) changes sign. - **Behavior of \(w(x)\):** its slope at any point is \(F(x)\). When \(F(x) > 0\), \(w(x)\) is increasing; when \(F(x) < 0\), \(w(x)\) is decreasing. --- ### Summary: - By plotting these functions together, you'll see how the oscillations of \(f(x)\) influence the shape of \(F(x)\), and in turn, how the shape of \(w(x)\) relates to \(F(x)\). --- ### Additional tip: If you're using Desmos: - For \(f(x)\): ``` f(x) = 3 \sin x \cos x^2 ``` - For \(F(x)\): ``` F(x) = \int_0^x 3 \sin t \cos t^2 dt ``` - For \(w(x)\): ``` w(x) = \int_0^x F(t) dt ``` Ensure your calculator or tool supports integral notation. If not, you can approximate the integrals numerically or compute sample points. --- Let me know if you'd like me to provide sample code snippets or further guidance!

Certainly! Here are the final concise answers: **a.** \(F(x)\) is *not necessarily* non-negative because \(f(t) = 3 \sin t \cos t^2\) can take negative values. **b.** The points where \(f(x) = 0\) correspond to the extrema of \(F(x)\), since \(F'(x) = f(x)\). **c.** Not every zero of \(f(x)\) corresponds to an extremum of \(F(x)\); only those where \(f(x)\) changes sign do. **d.** The function \(w(x) = \int_0^x F(t) dt\) is an *antiderivative* of \(F(x)\), so \(w'(x) = F(x)\), and \(w''(x) = f(x)\).