Application of Derivatives — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSApplication of Derivatives2 Marks
Question
Prove that the function $f$ given by $f(x) = x^2 - x + 1$ is neither strictly increasing nor decreasing on $(-1, 1)$.
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Answer
It is given that function $f(x) = x^2 - x + 1$
$f\ ' (x) = 2x - 1$
If $f\ ' (x) = 0$, then we get
$~~~ ~~~x=\frac{1}{2}$
So, the point $x = \frac{1}{2}$ divides the interval $(-1, 1)$ into two disjoint intervals, $\left(-1, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, 1\right)$
In interval $\left(-1, \frac{1}{2}\right)$, we have
$f\ ' (x) = 2x - 1 < 0$
Therefore, the given function $'f\ ' $ is strictly decreasing in interval $\left(-1, \frac{1}{2}\right)$
Also, in interval $\left(\frac{1}{2}, 1\right)$
$f\ ' (x) = 2x - 1 > 0$
Hence, the given function $'f\ ' $ is strictly increasing in interval for $\left(\frac{1}{2}, 1\right)$.
Therefore $, f$ is neither strictly increasing nor decreasing in interval $(-1, 1).$
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