Given the function \(f(x) = 2x^{3} - 5x^{2} + 6x - 3\), find the bounds of the zeros of this function.
Answer
Step 4: Use the above information to find the bounds of the zeros. Since \(f(x)\) is decreasing for \(x < \frac{1}{2}\) and increasing for \(\frac{1}{2} < x < 2\), there must be a zero in the interval \(x < \frac{1}{2}\). Similarly, since \(f(x)\) is increasing for \(\frac{1}{2} < x < 2\) and decreasing for \(x > 2\), there must be a zero in the interval \(x > 2\).
Step 1 ：Step 1: First, we find the derivative of the function. \(f'(x) = 6x^{2} - 10x + 6\)
Step 2 ：Step 2: Set the derivative equal to zero and solve for x to find critical points. \[6x^{2} - 10x + 6 = 0\] Solve this quadratic equation to get \[x = \frac{1}{2}, 2\]
Step 3 ：Step 3: Test the intervals determined by the critical points in the derivative. For \(x < \frac{1}{2}\), \(f'(x) < 0\), which means \(f(x)\) is decreasing. For \(\frac{1}{2} < x < 2\), \(f'(x) > 0\), which means \(f(x)\) is increasing. For \(x > 2\), \(f'(x) < 0\), which means \(f(x)\) is decreasing.
Step 4 ：Step 4: Use the above information to find the bounds of the zeros. Since \(f(x)\) is decreasing for \(x < \frac{1}{2}\) and increasing for \(\frac{1}{2} < x < 2\), there must be a zero in the interval \(x < \frac{1}{2}\). Similarly, since \(f(x)\) is increasing for \(\frac{1}{2} < x < 2\) and decreasing for \(x > 2\), there must be a zero in the interval \(x > 2\).
