GATE 2022 IN Q.28: solution and answer

GATE 2022 · INQ.28 · 1 mark · NAT

The global minimum of $x^{3} e^{- ∣ x ∣}$ for $x \in (- \infty, \infty)$ occurs at $x =$ ________ (round off to one decimal place)

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Answer and solution

ANSWER

The global minimum of $x^{3} e^{- ∣ x ∣}$ occurs at $x = - 3.0$ , where the function equals $- 27 e^{- 3} \approx - 1.344$ .

Official answer: -3.0Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Let $f (x) = x^{3} e^{- ∣ x ∣}$ . For $x \geq 0$ , both $x^{3} \geq 0$ and $e^{- ∣ x ∣} > 0$ , so $f (x) \geq 0$ . For $x < 0$ , $x^{3} < 0$ , so $f (x) < 0$ . The global minimum is therefore negative and must lie on $x < 0$ .
For $x < 0$ , $∣ x ∣ = - x$ , so $f (x) = x^{3} e^{x}$ .
Differentiate with the product rule: $f^{'} (x) = 3 x^{2} e^{x} + x^{3} e^{x} = x^{2} e^{x} (x + 3) .$
Critical points on $x < 0$ : there $x^{2} e^{x} > 0$ , so $f^{'} (x) = 0$ only when $x + 3 = 0$ , i.e. $x = - 3$ .
Sign of $f^{'}$ : for $x < - 3$ , $x + 3 < 0$ so $f^{'} (x) < 0$ and $f$ is decreasing. For $- 3 < x < 0$ , $f^{'} (x) > 0$ and $f$ is increasing. So on the whole half-line $x < 0$ , $f (x) \geq f (- 3)$ , with equality only at $x = - 3$ .
Value there: $f (- 3) = (- 3)^{3} e^{- 3} = - 27 e^{- 3} \approx - 1.344$ .
Global check: for every $x \geq 0$ , $f (x) \geq 0 > - 27 e^{- 3}$ . Together with step 5, $f (x) \geq - 27 e^{- 3}$ for all real $x$ . (As a sanity check, $f (x) \to 0$ as $x \to \pm \infty$ because the exponential decays faster than $x^{3}$ grows.) The global minimum therefore occurs at $x = - 3.0$ .

KEY IDEA

Remove the modulus by splitting at $x = 0$ . The function is non-negative for $x \geq 0$ , so the minimum must be on $x < 0$ , where $f (x) = x^{3} e^{x}$ . There $f^{'} (x) = x^{2} e^{x} (x + 3)$ changes sign from negative to positive only at $x = - 3$ , so $f$ decreases down to $x = - 3$ and rises after it. That monotonicity makes $x = - 3$ the global minimum, not just a local one.

WATCH OUT FOR

Writing $f (x) = x^{3} e^{- x}$ on the negative side (forgetting that $∣ x ∣ = - x$ there). That gives $f^{'} (x) = x^{2} e^{- x} (3 - x)$ and the answer $x = 3$ , which is actually where the global maximum occurs. Another slip is to report $x = 0$ from $f^{'} (x) = 0$ : $x^{2} = 0$ there, but $f^{'}$ does not change sign at $0$ , so $x = 0$ is neither a maximum nor a minimum.

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GATE 2022 IN Q.28: solution and answer

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