Question

2. \( f(x)=x^{2} \sqrt{x+1} \)

Ask by Willis Evans. in Canada
Jan 30,2025

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Answer

The derivative of \( f(x) = x^{2} \sqrt{x + 1} \) is \( f'(x) = \frac{x(5x + 4)}{2 \sqrt{x + 1}} \).

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The Deep Dive

The function \( f(x) = x^2 \sqrt{x + 1} \) showcases how polynomials combine with radical functions to create intriguing behaviors. As \( x \) becomes large, the function predominantly resembles \( x^{2.5} \) due to the square root, which causes rapid growth. However, the behavior changes dramatically as \( x \) approaches negative values because the square root becomes imaginary for \( x < -1 \), emphasizing the importance of the domain in function analysis. When evaluating or graphing this function, remember to consider its domain carefully. The radical requires \( x + 1 \geq 0\), meaning \( x \geq -1 \). A common mistake is to overlook this restriction, potentially leading to confusion about where the function is defined. Always check for values that might make a radical or denominator undefined—it's crucial for accurate results!

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