The Shape of a Graph, Part I — Question 6

Question 6

Problem:

Let $$f(x) = \frac{x^2}{x + 1}$$

• Find the critical points of $f(x)$ and determine where $f(x)$ is increasing or decreasing.

• Identify any local maximum or minimum values.

Question 6 - Solution

We are given: $$f(x) = \frac{x^2}{x + 1}$$

(a) First Derivative:

Use the quotient rule: $$f'(x) = \frac{(2x)(x + 1) - x^2(1)}{(x + 1)^2} = \frac{2x(x + 1) - x^2}{(x + 1)^2} = \frac{2x^2 + 2x - x^2}{(x + 1)^2} = \frac{x^2 + 2x}{(x + 1)^2} = \frac{x(x + 2)}{(x + 1)^2}$$

Set $f'(x) = 0$: $$\frac{x(x + 2)}{(x + 1)^2} = 0 \Rightarrow x = 0,\, -2$$

Also, note $f(x)$ is undefined at $x = -1$

Test intervals:

$(-\infty, -2)$: pick $x = -3$: numerator is positive, denominator is positive → $f' > 0$

$(-2, -1)$: pick $x = -1.5$: numerator is negative, denominator is positive → $f' < 0$

$(-1, 0)$: pick $x = -0.5$: numerator is negative, denominator is positive → $f' < 0$

$(0, \infty)$: pick $x = 1$: numerator is positive, denominator is positive → $f' > 0$

Conclusion: $$\begin{aligned} &f(x) \text{ increases on } (-\infty, -2) \cup (0, \infty) \\ &f(x) \text{ decreases on } (-2, -1) \cup (-1, 0) \end{aligned}$$

(b) Local Extrema:

At $x = -2$: from increasing to decreasing → local max $f(-2) = \frac{(-2)^2}{-2 + 1} = \frac{4}{-1} = -4$

At $x = 0$: from decreasing to increasing → local min $f(0) = 0$

Answer: $$\text{Local maximum at } (-2, -4), \quad \text{Local minimum at } (0, 0)$$

Graph of $f(x) = \frac{x^2}{x + 1}$: