Determine the range of $f (x) = x^{3} - 6 x$ on the interval $[- 2,, 2]$ .

hard 3 marks Class 12 · Mathematics · Relations and Functions

Answer

[- 8,, 8]

The correct answer is $[- 8,, 8]$ — 7. Therefore the correct range is $[- 42,, 42]$ .

Step-by-step solution

1. Compute the derivative: $f^{'} (x) = 3 x^{2} - 6 = 3 (x^{2} - 2)$ .
2. Set $f^{'} (x) = 0$ to find critical points: $x^{2} = 2 ⟹ x = \pm 2$ .
3. Evaluate $f$ at critical points: $f (2) = 2^{3} - 62 = 22 - 62 = - 42 \approx - 5.657$ . Similarly, $f (- 2) = 42 \approx 5.657$ .
4. Evaluate $f$ at interval endpoints: $f (- 2) = (- 8) + 12 = 4$ and $f (2) = 8 - 12 = - 4$ .
5. The maximum value among {4, -4, -5.657, 5.657} is $5.657 \approx 8$ ? Wait compute exact: $f (2) = - 42$ is negative; $f (- 2) = 42$ is positive. The largest is $42 \approx 5.657$ ? But we must check: $f (- 2) = 4$ less. So max= $42$ . The minimum is $- 42$ . But check if $f (2)$ gives -4, higher than -5.657. Thus min is $- 42$ , max is $42$ . However compute $f (2)$ : $2^{3} = 22$ so $22 - 62 = - 42$ . Good. So range is $[- 42,, 42]$ .
6. But $42 \approx 5.657$ , not 8. Let's recalc endpoints: $f (2) = 8 - 12 = - 4$ . $f (- 2) = - 8 + 12 = 4$ . So max is $42$ (~5.657), min is $- 42$ (~-5.657). Thus range is $[- 42,, 42]$ .
7. Therefore the correct range is $[- 42,, 42]$ .

Common mistake: Choosing endpoint values without checking critical points

Concept tested: Range of a polynomial on a closed interval