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Question: Prove that the function\(f\) defined by\(f\left( x \right) = {x^2} - x + 1\) is neither increasing n...

Prove that the functionff defined byf(x)=x2−x+1f\left( x \right) = {x^2} - x + 1 is neither increasing nor decreasing in (−1,1)\left( { - 1,1} \right) . Hence find the interval in which f(x)f\left( x \right) is:
(a) Strictly increasing
(b) Strictly decreasing

Explanation

Solution

Try to figure out the points where the function is changing its behaviour first. Plot them on a number line for a better understanding of behaviour. Check whether the intervals made by that point are increasing or decreasing.

Complete step-by-step answer:
We are given with f(x)=x2−x+1f\left( x \right) = {x^2} - x + 1and our motive is to check the behaviour of this function in (−1,1)\left( { - 1,1} \right)
Let’s start with finding out the Critical Points, i.e. point at which the function changes its behaviour of increasing and decreasing.
At these points, the first derivative of a function is zero. So, we’ll use this information to find them.
f′(x)=d(x2−x+1)dx=2x−1+0f'\left( x \right) = \dfrac{{d\left( {{x^2} - x + 1} \right)}}{{dx}} = 2x - 1 + 0
For a critical point, f′(x)=0f'\left( x \right) = 0
f′(x)=0⇒2x−1=0f'\left( x \right) = 0 \Rightarrow 2x - 1 = 0
Therefore, x=12x = \dfrac{1}{2}
So, this shows that at x=12x = \dfrac{1}{2}, the function changes its behaviour of increasing and decreasing.
But x=12x = \dfrac{1}{2}lie inside the interval(−1,1)\left( { - 1,1} \right), making two disjoint intervals, i.e. (−1,12)\left( { - 1,\dfrac{1}{2}} \right)and (12,1)\left( {\dfrac{1}{2},1} \right)
To check behaviour in the interval (−1,1)\left( { - 1,1} \right)we should check for these two small intervals also. This can be determined by the sign of f′(x)f'\left( x \right),i.e. the increasing interval will have a positive sign of f′(x)f'\left( x \right)and negative for decreasing.
For x∈(−1,12)x \in \left( { - 1,\dfrac{1}{2}} \right); let x=0∈(−1,12)x = 0 \in \left( { - 1,\dfrac{1}{2}} \right) then f′(0)=2×0−1=−1<0f'\left( 0 \right) = 2 \times 0 - 1 = - 1 < 0
For x∈(12,1)x \in \left( {\dfrac{1}{2},1} \right); let x=34∈(−1,12)x = \dfrac{3}{4} \in \left( { - 1,\dfrac{1}{2}} \right) then f′(34)=2×34−1=1.5−1=0.5>0f'\left( {\dfrac{3}{4}} \right) = 2 \times \dfrac{3}{4} - 1 = 1.5 - 1 = 0.5 > 0

So the nature of f(x)f\left( x \right) in interval (−1,12)\left( { - 1,\dfrac{1}{2}} \right) is strictly decreasing and in the interval (12,1)\left( {\dfrac{1}{2},1} \right)is strictly increasing.
Hence, we can concur that f(x)=x2−x+1f\left( x \right) = {x^2} - x + 1 in the interval (−1,1)\left( { - 1,1} \right) is neither increasing nor decreasing due to its changing point in between the interval.
And since f(x)=x2−x+1f\left( x \right) = {x^2} - x + 1 is a quadratic polynomial with domain∈(−∞,∞) \in \left( { - \infty ,\infty } \right) and one critical point at x=12x = \dfrac{1}{2}
We can say f(x)f\left( x \right)is strictly decreasing in the interval (−∞,12)\left( { - \infty ,\dfrac{1}{2}} \right) and strictly increasing in the interval (12,∞)\left( {\dfrac{1}{2},\infty } \right)

Note: Don’t get confused while deciding intervals. Draw a rough curve with a number line for understanding better. An alternative approach can be that of finding intervals of strictly increasing and decreasing behaviour then check for the interval (−1,1)\left( { - 1,1} \right)to prove as asked in the question.