The given function is \( f(x) = x^3 - 3x^2 - 4x \).
To find the critical points, we need to find the first derivative of the function and set it to zero.
\[ f'(x) = 3x^2 - 6x - 4 \]
Set the derivative equal to zero:
\[ 3x^2 - 6x - 4 = 0 \]
Solve the quadratic equation:
\[ 3x^2 - 6x - 4 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{6 \pm \sqrt{36 + 48}}{6} \]
\[ x = \frac{6 \pm \sqrt{84}}{6} \]
\[ x = \frac{6 \pm 2\sqrt{21}}{6} \]
\[ x = 1 \pm \frac{\sqrt{21}}{3} \]
Evaluate the second derivative to determine concavity:
\[ f''(x) = 6x - 6 \]
At \( x = 1 + \frac{\sqrt{21}}{3} \):
\[ f''\left(1 + \frac{\sqrt{21}}{3}\right) = 6\left(1 + \frac{\sqrt{21}}{3}\right) - 6 = 2\sqrt{21} > 0 \]
This indicates a local minimum.
At \( x = 1 - \frac{\sqrt{21}}{3} \):
\[ f''\left(1 - \frac{\sqrt{21}}{3}\right) = 6\left(1 - \frac{\sqrt{21}}{3}\right) - 6 = -2\sqrt{21} < 0 \]
This indicates a local maximum.
As \( x \to \infty \), \( f(x) \to \infty \).
As \( x \to -\infty \), \( f(x) \to -\infty \).
The graph should have a local maximum and minimum, and the end behavior should match the cubic function characteristics.
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