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Question 8
Suppose that a and b are positive real numbers, and let $f(x) = \frac{a + b + x}{3(ab)^{\frac{1}{3}}}$ for $x > 0$. (i) Show that the minimum value of $f(x)$ occurs... show full transcript
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Answer
To find the minimum value of the function , consider its derivative:
Since the derivative is constant and positive for , the function is monotonically increasing. Therefore, the minimum occurs at the left endpoint of the interval; let’s check the value at :
f\left(\frac{a + b}{2}\right) = \frac{a + b + \frac{a + b}{2}}{3(ab)^{\frac{1}{3}} = \frac{\frac{3(a + b)}{2}}{3(ab)^{\frac{1}{3}} = \frac{a + b}{2(ab)^{\frac{1}{3}}}.
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Answer
Using the discriminant condition derived from previous parts and the derived inequalities, we find the cubic's nature through critical points and intercept evaluations, concluding that it crosses the x-axis exactly once, indicating one real root.
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