Question

range of f(x) = x^3 - 12x without drawing graph

Answer 1

(-\infty, \infty)

Answer 2

To find the range of the function $f(x) = x^3 - 12x$, we use calculus to identify its local extrema and analyze its behavior as $x$ approaches positive and negative infinity.

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

2. Find critical points: Set $f'(x) = 0$ to find the critical points: $3x^2 - 12 = 0$ $3(x^2 - 4) = 0$ $x^2 - 4 = 0$ $(x - 2)(x + 2) = 0$ This gives two critical points: $x = 2$ and $x = -2$.

3. Determine the nature of critical points (local maxima/minima): We can use the second derivative test. $f''(x) = \frac{d}{dx}(3x^2 - 12) = 6x$.

   • At $x = 2$: $f''(2) = 6(2) = 12$. Since $f''(2) > 0$, there is a local minimum at $x = 2$. The value of the function at this local minimum is $f(2) = (2)^3 - 12(2) = 8 - 24 = -16$.

   • At $x = -2$: $f''(-2) = 6(-2) = -12$. Since $f''(-2) < 0$, there is a local maximum at $x = -2$. The value of the function at this local maximum is $f(-2) = (-2)^3 - 12(-2) = -8 + 24 = 16$.

4. Analyze the behavior as $x \to \pm \infty$: Since $f(x)$ is a polynomial of odd degree (degree 3) with a positive leading coefficient (1), its end behavior is:

   • As $x \to \infty$, $f(x) = x^3 - 12x \to \infty$.
   • As $x \to -\infty$, $f(x) = x^3 - 12x \to -\infty$.

5. Determine the range: The function starts from $-\infty$, increases to a local maximum of 16 at $x = -2$, then decreases to a local minimum of -16 at $x = 2$, and finally increases to $\infty$. Since the function is continuous and covers all values from $-\infty$ to $\infty$, its range is all real numbers.

The range of $f(x) = x^3 - 12x$ is $(-\infty, \infty)$.