Question

Let 𝑓(𝑥) = 1 3 𝑥 3 + 𝑥 2 − 15𝑥 − 9 . Use detailed sign tables in answering the following questions. (a) Find the intervals in which 𝑓 is increasing or decreasing. (b) Find the intervals in which the graph of 𝑦 = 𝑓(𝑥) is concave upwards or downwards. 

Answer 1

As I understood, the function is $f(x) = 13x^3+x^2-15x-9$ .

a) Then, intervals in which f(x) is increasing $\leftrightarrow$ $f'(x) > 0$ .

Intervals in which f(x) is decreasing $\leftrightarrow$$f'(x) < 0$ .

$f'(x) = 39x^2+2x-15 \longrightarrow$ find nulls of $f'(x)$

$x_{1,2} = \frac{1}{78}(-2 \pm \sqrt{4+4*15*39}) = \\ \frac{1}{39}(-1 \pm \sqrt{586}), x_1 < x_2$

We have found null of quadratic trinomial, so since it's "branches" are directed up :

$x < x_1 \longrightarrow f'(x) >0 \\ x_1 < x < x_2 \longrightarrow f'(x) <0 \\ x_2 < x \longrightarrow f'(x) >0$

So, the answer :

f(x) increases on $(-\infty; -\frac{1}{39} - \frac{\sqrt{586}}{39}) \: \cup \: (-\frac{1}{39} + \frac{\sqrt{586}}{39}; +\infty)$

f(x) decreases on $(-\frac{1}{39}-\frac{\sqrt{586}}{39};-\frac{1}{39} + \frac{\sqrt{586}}{39})$

b) Plot of $y = f(x)$ is concave upwards if $f''(x) < 0$ .

Plot of $y = f(x)$ is concave downwards if $f''(x) > 0$

$f''(x) = (f'(x))' = 78x+2, f''(x_0) = 0 \longrightarrow x_0 = -\frac{1}{39}$

So, ... upwards if $(x < -\frac{1}{39})$ , and ... downwards if $(-\frac{1}{39} < x)$