Question:

Step 1: Expand the function and introduce a substitution. We begin by expanding the given function: $$f(x) = \cos x(1 + \cos x) = \cos x + \cos^2 x$$ Let $t = \cos x$. Since the range of cosine is $[-1, 1]$, we have $t \in [-1, 1]$. This transforms our problem into finding the maximum and minimum of: $$g(t) = t + t^2 \text{ where } t \in [-1, 1]$$ Step 2: Complete the square to identify the vertex form. We rewrite $g(t)$ by completing the square: $$g(t) = t^2 + t = \left(t + \frac{1}{2}\right)^2 - \frac{1}{4}$$ This is a parabola opening upward (since the coefficient of $t^2$ is positive) with vertex at $t = -\frac{1}{2}$. Step 3: Find the minimum value on the interval $[-1, 1]$. Since the vertex is at $t = -\frac{1}{2}$ and this point lies within our domain $[-1, 1]$, the minimum value occurs at the vertex: $$g\left(-\frac{1}{2}\right) = -\frac{1}{2} + \left(-\frac{1}{2}\right)^2 = -\frac{1}{2} + \frac{1}{4} = -\frac{1}{4}$$ Step 4: Find the maximum value on the interval $[-1, 1]$. For a parabola opening upward, the maximum on a closed interval occurs at one of the endpoints. We evaluate: - At $t = -1$: $g(-1) = -1 + 1 = 0$ - At $t = 1$: $g(1) = 1 + 1 = 2$ The maximum value is $g(1) = 2$. Step 5: Calculate the difference between maximum and minimum. The difference between the maximum and minimum values is: $$\text{Maximum} - \text{Minimum} = 2 - \left(-\frac{1}{4}\right) = 2 + \frac{1}{4} = \frac{9}{4}$$ **Final Answer:** The maximum value of $f(x)$ is greater than its minimum value by $\boxed{\frac{9}{4}}$, which corresponds to **Option 4**.