Question

For which values of $x$, the function $f\left( x \right)={{x}^{2}}-2x$ is decreasing

1. $x > 1$
2. $x > 2$
3. $x < 1$
4. $x < 2$

Answer 1

a function with a graph that moves downwards as it follows from left to right. if a function is differentiable, then it is decreasing at all points where its derivative is negative. If $f'\left( x \right)<0$ for every $x$ on some interval, then $f\left( x \right)$ is decreasing on the interval.

Complete step by step answer:
From the question it is clear that we have to find the values of $x$for which the function $f\left( x \right)={{x}^{2}}-2x$ is decreasing
To check whether the function is increasing or decreasing, first of all the function should be differentiable.
Graph moves downwards as it follows from left to right.
if a function is differentiable, then it is decreasing at all points where its derivative is negative. If $f'\left( x \right) < 0$ for every $x$ on some interval, then $f\left( x \right)$ is decreasing on the interval.
These are some properties of decreasing function.
Consider the equation from the given question,
$\Rightarrow f\left( x \right)={{x}^{2}}-2x$
Differentiate the given function with respect to $x$. So,
$\Rightarrow f'\left( x \right)=\dfrac{d}{dx}({{x}^{2}}-2x)$
$\Rightarrow f'\left( x \right)=2x-2$
For decreasing function $f'\left( x \right)<0$. So,
$\Rightarrow 2x-2 < 0$
Now add $2$ on both sides, we get
$\Rightarrow 2x-2+2 < 0+2$
On simplification we get,
$\Rightarrow 2x < 2$
$\Rightarrow \dfrac{2x}{2} < \dfrac{2}{2}$
$\Rightarrow x < 1$
Now we can conclude that any value less than $1$, gives a decreasing function.

So, the correct answer is “Option 3”.

Note: students should be careful while doing calculations because small calculation errors can make a large difference in the final answer. Many students may have the misconception that If $f'\left( x \right) > 0$ for every $x$ on some interval, then $f\left( x \right)$ is decreasing on the interval. But actually it is an increasing function for $f'\left( x \right) > 0$.