The maximum value of the function \( f(x) = (x - 1)(x - 2)(x - 3) \) in the domain [0, 3] occurs at \( x = \) _________ (rounded off to two decimal places).
The maximum value of the function \( f(x) = (x - 1)(x - 2)(x - 3) \) in the domain [0, 3] occurs at \( x = \) _________ (rounded off to two decimal places).
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Solution and Explanation
We are tasked with finding the maximum value of the function \( f(x) = (x - 1)(x - 2)(x - 3) \) in the domain \( [0, 3] \).
Step 1: Find the first derivative of the function to locate the critical points. We first differentiate \( f(x) \) using the product rule: \[ f'(x) = \frac{d}{dx} \left[ (x - 1)(x - 2)(x - 3) \right] \] To simplify the differentiation, expand the function first: \[ f(x) = (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6 \] Now, differentiate: \[ f'(x) = 3x^2 - 12x + 11 \] Step 2: Solve for the critical points by setting the derivative equal to zero. Set \( f'(x) = 0 \): \[ 3x^2 - 12x + 11 = 0 \] Solving this quadratic equation using the quadratic formula: \[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(11)}}{2(3)} = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} \] \[ x = \frac{12 \pm 2\sqrt{3}}{6} \] \[ x = 2 \pm \frac{\sqrt{3}}{3} \] The two critical points are approximately: \[ x \approx 2 + 0.577 = 2.58 \quad {and} \quad x \approx 2 - 0.577 = 1.41 \] Step 3: Evaluate the function at the critical points and endpoints.
Now, evaluate \( f(x) \) at the critical points \( x = 2.58 \), \( x = 1.41 \), and at the endpoints \( x = 0 \) and \( x = 3 \).
\( f(0) = (0 - 1)(0 - 2)(0 - 3) = (-1)(-2)(-3) = -6 \)
\( f(3) = (3 - 1)(3 - 2)(3 - 3) = (2)(1)(0) = 0 \)
\( f(1.41) = (1.41 - 1)(1.41 - 2)(1.41 - 3) = (0.41)(-0.59)(-1.59) \approx 0.384 \)
\( f(2.58) = (2.58 - 1)(2.58 - 2)(2.58 - 3) = (1.58)(0.58)(-0.42) \approx -0.384 \)
Step 4: Conclusion
The maximum value occurs at \( x = 1.41 \), and the value of the function is approximately \( 0.384 \), but rounded to two decimal places, the maximum occurs at: \[ \boxed{1.41} \]
Top Questions on Function
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