Question
Find the intervals in which the function $f $ given by $f(x) = 2x^2 – 3x$ is increasing.
✓
Answer
It is given that function $, f(x) = 2x^2 - 3x$
$\Rightarrow f\ ' (x) = 4x - 3$
If $f\ ' (x) = 0,$ then we get,
$\mathrm{x}=\frac{3}{4}$
So, the point $x = \frac{3}{4}$, divides the real line into two disjoint intervals, $\left(-\infty, \frac{3}{4}\right)$ and $\left(\frac{3}{4}, \infty\right)$
So, in interval $\left(\frac{3}{4}, \infty\right), f\ ' (x) = 4x -3 > 0$
Therefore, the given function $(f)$ is strictly increasing in interval $\left(\frac{3}{4}, \infty\right)$
$\Rightarrow f\ ' (x) = 4x - 3$
If $f\ ' (x) = 0,$ then we get,
$\mathrm{x}=\frac{3}{4}$
So, the point $x = \frac{3}{4}$, divides the real line into two disjoint intervals, $\left(-\infty, \frac{3}{4}\right)$ and $\left(\frac{3}{4}, \infty\right)$
So, in interval $\left(\frac{3}{4}, \infty\right), f\ ' (x) = 4x -3 > 0$
Therefore, the given function $(f)$ is strictly increasing in interval $\left(\frac{3}{4}, \infty\right)$
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