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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