Question
Find the interval of the function that is strictly increasing or decreasing: $10 - 6x - 2x^2$
✓
Answer
It is given that function $f(x) = 10 - 6x - 2x^2$
$f'(x) = -6 - 4x$
If f'(x) = 0, then we get,
$\Rightarrow$ $x=\frac{-3}{2}$
So, the point $x=\frac{-3}{2}$ divides the real line into two disjoint intervals, $\left(-\infty, \frac{-3}{2}\right)$ and $\left(\frac{-3}{2}, \infty\right)$
So, in interval $\left(-\infty, \frac{-3}{2}\right)$
x < $\frac{-3}{2}$
$\Rightarrow$ -4x > 6
$\Rightarrow$-6 - 4x > 0
i.e, f'(x) > 0
Therefore, the given function (f) is strictly increasing in interval $\left(-\infty, \frac{-3}{2}\right)$ and in interval $\left(\frac{-3}{2}, \infty\right)$
f'(x) = -6 - 4x < 0
Therefore, the given function (f) is strictly decreasing in interval $\left(\frac{-3}{2}, \infty\right)$
Thus, function is strictly increasing in $\left(-\infty, \frac{-3}{2}\right)$and strictly decreasing in $\left(\frac{-3}{2}, \infty\right)$
$f'(x) = -6 - 4x$
If f'(x) = 0, then we get,
$\Rightarrow$ $x=\frac{-3}{2}$
So, the point $x=\frac{-3}{2}$ divides the real line into two disjoint intervals, $\left(-\infty, \frac{-3}{2}\right)$ and $\left(\frac{-3}{2}, \infty\right)$
So, in interval $\left(-\infty, \frac{-3}{2}\right)$
x < $\frac{-3}{2}$
$\Rightarrow$ -4x > 6
$\Rightarrow$-6 - 4x > 0
i.e, f'(x) > 0
Therefore, the given function (f) is strictly increasing in interval $\left(-\infty, \frac{-3}{2}\right)$ and in interval $\left(\frac{-3}{2}, \infty\right)$
f'(x) = -6 - 4x < 0
Therefore, the given function (f) is strictly decreasing in interval $\left(\frac{-3}{2}, \infty\right)$
Thus, function is strictly increasing in $\left(-\infty, \frac{-3}{2}\right)$and strictly decreasing in $\left(\frac{-3}{2}, \infty\right)$
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