Study of a cubic, intercepts and extrema

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Consider the cubic function $f(x) = x^3 - x^2 - 56x$.

• (a) Find the intercepts with the axes.
• (b) Compute the derivative $f'(x)$ and study its sign to determine where $f$ increases and decreases.
• (c) Find the values of $x$ where the graph has a relative maximum or minimum, and the corresponding value of $f$.

Solution

(a) Intercepts. $f(x) = x(x^2 - x - 56) = x(x-8)(x+7)$, so with the $x$-axis: $x=-7,\ 0,\ 8$. With the $y$-axis: $f(0)=0$, point $(0,0)$.

(b) Derivative and monotonicity. $f'(x) = 3x^2 - 2x - 56$, which vanishes at $x = \frac{2 \pm \sqrt{4 + 672}}{6} = \frac{2 \pm 26}{6} \;\Rightarrow\; x=-4 \ \text{or}\ x=\tfrac{14}{3}.$ Since the parabola $f'$ opens upward, $f'>0$ (increasing) for $x<-4$ and $x>\tfrac{14}{3}$; $f'<0$ (decreasing) for $-4<x<\tfrac{14}{3}$.

(c) Extrema. At $x=-4$ the derivative changes from $+$ to $-$: relative maximum, with $f(-4) = -64 - 16 + 224 = 144.$ At $x=\tfrac{14}{3}$ the derivative changes from $-$ to $+$: relative minimum, with $f\!\left(\tfrac{14}{3}\right) = -\tfrac{4900}{27} \approx -181.48.$ $\boxed{\text{max } (-4,\,144);\quad \text{min } \left(\tfrac{14}{3},\,-\tfrac{4900}{27}\right)}$