Find the two \( x \)-intercepts of the function \( f \) and show that \( f^{\prime}(x)=0 \) at some point between the two \( x \)-intercepts. \[ f(x)=x(x-6) \] smaller \( x \)-value \( \quad(x, y)= \) \( \square \) ) larger \( x \)-value \( \quad(x, y)= \) \( \square \) Find a value of \( x \) such that \( f^{\prime}(x)=0 \). \[ x=\square \] Need Help? Road It Waich It

To find the \( x \)-intercepts of the function \( f(x) = x(x-6) \), we set \( f(x) = 0 \): \[ x(x - 6) = 0 \] This gives us two solutions: 1. \( x = 0 \) 2. \( x - 6 = 0 \) which simplifies to \( x = 6 \) Thus, the smaller \( x \)-value is \( (0, 0) \) and the larger \( x \)-value is \( (6, 0) \). Now we find the derivative \( f^{\prime}(x) \): \[ f^{\prime}(x) = 1 \cdot (x-6) + x \cdot 1 = 2x - 6 \] Setting the derivative equal to zero to find the critical points: \[ 2x - 6 = 0 \quad \Rightarrow \quad 2x = 6 \quad \Rightarrow \quad x = 3 \] Since \( x = 3 \) is indeed between the \( x \)-intercepts 0 and 6, we have completed the task! Smaller \( x \)-value \( \quad(x, y)= \) \( (0, 0) \) Larger \( x \)-value \( \quad(x, y)= \) \( (6, 0) \) Find a value of \( x \) such that \( f^{\prime}(x)=0 \). \( x=\) \( 3 \) Here's a fun fact: the quadratic \( f(x) = x(x-6) \) opens upwards, meaning its \( x \)-intercepts represent the roots of the equation, and the vertex lies right in between them. So where do we find that critical point? Right at \( x = 3 \) — the peak performance of our parabola! And for some problem-solving tips: always double-check whether your factors yield the correct roots! Sometimes signs can be tricky. In this case, factoring easily gave the intercepts. Just remember, if you ever get stuck, look for alternative methods like using the quadratic formula as a backup plan!