Question

find range of f(x) = x^3 - 12x if x belongs to [-3,1]

Answer 1

[-11, 16]

Answer 2

To find the range of $f(x) = x^3 - 12x$ for $x \in [-3, 1]$, we need to find the absolute maximum and minimum values of the function on this closed interval.

1. Find the derivative of the function: $f(x) = x^3 - 12x$ $f'(x) = \frac{d}{dx}(x^3 - 12x) = 3x^2 - 12$

2. Find the critical points: Set $f'(x) = 0$: $3x^2 - 12 = 0$ $3(x^2 - 4) = 0$ $x^2 - 4 = 0$ $(x - 2)(x + 2) = 0$ The critical points are $x = 2$ and $x = -2$.

3. Check which critical points lie within the given interval: The given interval is $[-3, 1]$.

   • $x = 2$: This point is not in $[-3, 1]$.
   • $x = -2$: This point is in $[-3, 1]$.

4. Evaluate the function at the critical points within the interval and at the endpoints of the interval:

   • At the critical point $x = -2$: $f(-2) = (-2)^3 - 12(-2) = -8 + 24 = 16$
   • At the left endpoint $x = -3$: $f(-3) = (-3)^3 - 12(-3) = -27 + 36 = 9$
   • At the right endpoint $x = 1$: $f(1) = (1)^3 - 12(1) = 1 - 12 = -11$

5. Determine the range: The values of the function at these points are $16$, $9$, and $-11$. The absolute minimum value is the smallest of these values, which is $-11$. The absolute maximum value is the largest of these values, which is $16$. Since the function is continuous, it takes on all values between its absolute minimum and absolute maximum on the interval.

Therefore, the range of $f(x)$ for $x \in [-3, 1]$ is $[-11, 16]$.