
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
509.7k+ views
Hint: 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 \[f'\left( x \right) > 0\].
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 \[f'\left( x \right) > 0\].
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