The Shape of a Graph, Part I — Question 3

Question 3 - Solution

We are given $$f(x) = \frac{x^2 - 4x}{x + 1} = \frac{x(x-4)}{x+1}, \qquad x \neq -1$$

(a) First derivative

Using the quotient rule, $$f'(x) = \frac{(x+1)(2x-4)-(x^2-4x)}{(x+1)^2} = \frac{x^2+2x-4}{(x+1)^2}.$$

Setting the numerator equal to zero gives $$x^2+2x-4=0 \Rightarrow x=-1\pm\sqrt{5}.$$

Since $(x+1)^2>0$ for $x\neq -1$, the sign of $f'(x)$ is determined by the numerator.

Answer

The function is increasing on $$\boxed{(-\infty,-1-\sqrt{5}) \cup (-1+\sqrt{5},\infty)}.$$

The function is decreasing on $$\boxed{(-1-\sqrt{5},-1) \cup (-1,-1+\sqrt{5})}.$$

(b) Local extrema

At $x=-1-\sqrt{5}$, the derivative changes from positive to negative, so there is a local maximum.

At $x=-1+\sqrt{5}$, the derivative changes from negative to positive, so there is a local minimum.

$$f(-1-\sqrt{5})=-6-2\sqrt{5}, \qquad f(-1+\sqrt{5})=-6+2\sqrt{5}.$$

Answer

Local maximum at $$\boxed{\left(-1-\sqrt{5},-6-2\sqrt{5}\right)}$$

Local minimum at $$\boxed{\left(-1+\sqrt{5},-6+2\sqrt{5}\right)}$$

Graph of $f(x)$