Question

Let I be an open interval contained in the domain of a real function f, then $ f\left( x \right) $ is called strictly increasing function in I if
A. $ {{x}_{1}}>{{x}_{2}} $ in I gives \[f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right)\]
B. $ {{x}_{1}}<{{x}_{2}} $ in I gives \[f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right)\]
C. $ {{x}_{1}}={{x}_{2}} $ in I gives \[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
D. $ {{x}_{1}}={{x}_{2}} $ in I gives \[f\left( {{x}_{1}} \right)

Answer 1

Hint: We first try to describe the relation between the slope of a curve and the characteristics of it being increasing, decreasing. We find the differentiation of the curve for $ y=f\left( x \right) $ . Depending on the value of slope we get the characteristics of the function.

Complete step by step solution:
We first try to find the general term of a function where $ y=f\left( x \right) $ . We express the terms as $ {{t}_{n}} $ , the $ {{n}^{th}} $ term of the series.
We take differentiation of the function and find the slope of the function.
So, $ \dfrac{df}{dx}={{f}^{'}}\left( x \right) $ is the slope of the function.
Now, if the slope at any fixed point is negative which means $ \dfrac{df}{dx}<0 $ then the function is decreasing and if $ \dfrac{df}{dx}>0 $ then the function is increasing.
If the changes for the whole curve happens very rapidly then the function is not monotone.
We can also find the value of $ x $ for which if we get $ {{x}_{1}}>{{x}_{2}} $ and $ f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right) $ , the curve is increasing. If we find $ {{x}_{1}}<{{x}_{2}} $ and $ f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right) $ , the curve is decreasing. The change of values is equal to the slope. The correct option is A.
So, the correct answer is “Option A”.

Note: Let’s take as an example where $ f\left( x \right)=2x $ .
We find the slope of the function by taking $ \dfrac{df}{dx}={{f}^{'}}\left( x \right) $ .
So, $ \dfrac{df}{dx}={{f}^{'}}\left( x \right)=2 $ as \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\].
Now for any value of $ x $ , the value of $ \dfrac{df}{dx}={{f}^{'}}\left( x \right)=2>0 $ .
The function is monotonically increasing the whole function.
If we change the function from $ f\left( x \right)=2x $ to $ f\left( x \right)=-2x $ , the function becomes monotonically decreasing as $ \dfrac{df}{dx}=\dfrac{d}{dx}\left( -2x \right)=-2<0 $ for any value of $ x $ .