Question

let $f(x)=6x^{4}+2x^{3}$. select all intervals where $f(x)$ is increasing,
$(-\frac{1}{4},0)$
$(-\infty,\frac{-1}{3})$
$(0,\infty)$
$(\frac{-1}{3},\frac{-1}{4})$

Explanation:

Step1: Find the derivative

The derivative of \(f(x)=6x^{4}+2x^{3}\) is \(f^\prime(x)=24x^{3}+6x^{2}=6x^{2}(4x + 1)\)

Step2: Determine where \(f^\prime(x)>0\)

Since \(6x^{2}\geq0\) for all \(x\in R\), the sign of \(f^\prime(x)\) is determined by \(4x + 1\).
Set \(4x+1>0\), we get \(x>-\frac{1}{4}\).
When \(x = 0\), \(f^\prime(0)=0\). For \(x>0\), \(f^\prime(x)=6x^{2}(4x + 1)>0\) (because \(x^{2}>0\) and \(4x + 1>1\)).
For \(x\in(-\frac{1}{4},0)\), \(x^{2}>0\) and \(4x + 1>0\), so \(f^\prime(x)>0\)

Answer:

\((-\frac{1}{4},0)\), \((0,\infty)\)