Question

The function f(x)=x⁵e^-x is increasing in the interval

• A. (5,∞)
• B. (4,∞)
• C. (-4,∞)
• D. (-∞,5)
• E. (-5,∞)

Answer 1

The Correct Option is D

We are given the function \( f(x) = x^5 e^{-x} \). To find the interval where the function is increasing, we first calculate the derivative of the function: \[ f'(x) = \frac{d}{dx} \left( x^5 e^{-x} \right) \] We apply the product rule to differentiate: \[ f'(x) = \frac{d}{dx} (x^5) \cdot e^{-x} + x^5 \cdot \frac{d}{dx} (e^{-x}) \] \[ f'(x) = 5x^4 e^{-x} - x^5 e^{-x} \] Now, factor out \( e^{-x} \) (since it is always positive): \[ f'(x) = e^{-x} \left( 5x^4 - x^5 \right) \] \[ f'(x) = e^{-x} x^4 (5 - x) \] For \( f'(x) > 0 \), the term \( x^4 (5 - x) \) must be positive. Since \( x^4 \) is always positive, we require \( 5 - x > 0 \), which gives: \[ x < 5 \] Thus, the function is increasing in the interval \( (-\infty, 5) \).

The correct option is (D) : \((-∞,5)\)

Answer 2

We are given the function \(f(x) = x^5 e^{-x}\). To find the interval where the function is increasing, we need to find where its derivative, \(f'(x)\), is positive.

First, let's find the derivative of \(f(x)\) using the product rule: \[f'(x) = \frac{d}{dx}(x^5 e^{-x}) = \frac{d}{dx}(x^5) \cdot e^{-x} + x^5 \cdot \frac{d}{dx}(e^{-x})\] \[f'(x) = 5x^4 e^{-x} + x^5 (-e^{-x}) = 5x^4 e^{-x} - x^5 e^{-x}\]

Factor out \(x^4 e^{-x}\): \[f'(x) = x^4 e^{-x}(5 - x)\]

Now, we want to find where \(f'(x) > 0\). Since \(e^{-x}\) is always positive for any real number \(x\), and \(x^4\) is non-negative, the sign of \(f'(x)\) depends on the sign of \((5 - x)\).

We have \(f'(x) > 0\) when \(5 - x > 0\), which means \(x < 5\). Also, when x = 0, f'(x) = 0. Also x^4 = 0 when x = 0, so f'(0) = 0.

Since \(x^4\) is zero at x=0, we also need to examine what happens around x=0. If x is slightly less than 0, then \(x^4\) is positive and 5-x is positive, so f'(x) is positive. Similarly if x is slightly more than 0, then \(x^4\) is positive and 5-x is positive, so f'(x) is positive. Thus, f(x) is increasing for all x<5.

Therefore, the function \(f(x)\) is increasing in the interval \((-\infty, 5)\).