If a_n is the greatest term in the sequence a | vedclass

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(B) Let $f(x) = \frac{x^3}{x^4+147}$.To find the maximum value,we calculate the derivative $f'(x)$:$f'(x) = \frac{(x^4+147)(3x^2) - (x^3)(4x^3)}{(x^4+

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$5$

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(B) Let $f(x) = \frac{x^3}{x^4+147}$.To find the maximum value,we calculate the derivative $f'(x)$:$f'(x) = \frac{(x^4+147)(3x^2) - (x^3)(4x^3)}{(x^4+147)^2}$$f'(x) = \frac{3x^6 + 441x^2 - 4x^6}{(x^4+147)^2} = \frac{441x^2 - x^6}{(x^4+147)^2} = \frac{x^2(441 - x^4)}{(x^4+147)^2}$.Setting $f'(x) = 0$,we get $x^2 = 0$ or $x^4 = 441$,which implies $x^2 = 21$ (since $x > 0$),so $x = \sqrt{21} \approx 4.58$.Since $f(x)$ increases for $x  \sqrt{21}$,the maximum term in the sequence $a_n$ occurs at $n$ values closest to $\sqrt{21}$.We compare $a_4$ and $a_5$:$a_4 = \frac{4^3}{4^4+147} = \frac{64}{256+147} = \frac{64}{403} \approx 0.1588$.$a_5 = \frac{5^3}{5^4+147} = \frac{125}{625+147} = \frac{125}{772} \approx 0.1619$.Since $a_5 > a_4$,the greatest term is at $n = 5$.

If $a_n$ is the greatest term in the sequence $a_n = \frac{n^3}{n^4+147}$,$n = 1, 2, 3, \ldots$,then $n$ is equal to $..........$.

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