The Shape of a Graph, Part I — Question 5

Question 5 - Solution

We are given: $$f(x) = x^4 - 4x^3 + 6$$

(a) First Derivative: $$f'(x) = 4x^3 - 12x^2$$ $$f'(x) = 4x^2(x - 3)$$

Set $f'(x) = 0$: $$4x^2(x - 3) = 0 \quad \Rightarrow \quad x = 0,\ 3$$

Test intervals:

For $x < 0$: pick $x = -1$: $f'(-1) = 4(-1)^2(-1 - 3) = 4(1)(-4) = -16 < 0$

For $0 < x < 3$: pick $x = 1$: $f'(1) = 4(1)^2(1 - 3) = 4(-2) = -8 < 0$

For $x > 3$: pick $x = 4$: $f'(4) = 4(16)(1) = 64 > 0$

Conclusion: $$\begin{aligned} &f(x) \text{ is decreasing on } (-\infty, 3) \\ &f(x) \text{ is increasing on } (3, \infty) \end{aligned}$$

(b) Local Extrema:

$f'(x)$ does not change sign around $x = 0$ (negative on both sides) → no extremum at $x = 0$ At $x = 3$: from decreasing to increasing → local minimum

Evaluate: $$f(3) = 3^4 - 4(3)^3 + 6 = 81 - 108 + 6 = -21$$

Local Minimum: $\boxed{(3,\ -21)}$

Graph of $f(x) = x^4 - 4x^3 + 6$: