Question

For which region is $f\left(x\right)=3x^2-2x+1$ strictly increasing?

• A. $(2,5)$
• B. $(\frac{1}{3},\infty )$
• C. $(-1,\frac{1}{3}]$
• D. $(-\infty ,\frac{1}{3})$

Answer 1

Answer: (A), (B)

$(2,5)$
$(\frac{1}{3},\infty )$

To find where the function is increasing

1) Find its derivatives

Given : $f\left(x\right)=3x^2-2x+1$

$f^1\left(x\right)=6x-2$

Equate it to zero

$6x-2=0$

$6x=2$

$x=\frac{1}{3}$ (So the values which make derivatives equal to $0$ is $\frac{1}{3}$)

$\to$ Split and separate the value between intervals $(-\infty ,\infty )$

$\to$ Which gives

$(-\infty ,\frac{1}{3})(\frac{1}{3},\infty ,)$
Since $(2,5)$ and $(\frac{1}{3},\infty )$ falls under this intervals.
Hense both options A and B is the correct answers