Question

In the interval \[\left[ {0,1} \right]\] the function \[{x^2} - x + 1\] is
A.Increasing
B.Decreasing
C.Neither increasing nor decreasing
D.None of the above

Answer 1

Hint: Firstly find the derivative of a function as it can be used to determine if the function is increasing or decreasing on an interval in its domain. If \[f'(x) > 0\] at each point in an interval I, then the function is said to be increasing on I. if \[f'(x) < 0\] at each point in an interval I, then the function is said to be decreasing on I.

Complete answer:
If \[f'(x) > 0\] then \[f\] is increasing on the interval, and if \[f'(x) < 0\] then \[f\] is decreasing on the interval.
Following steps are involved in the process of finding the intervals of increasing and decreasing function:
Firstly, differentiate the given function with respect to the constant variable.
Then solve \[f'(x) = 0\] .
After solving the equation of the first derivative and finding the points of discontinuity we get the open intervals with the value of \[x\], through which the sign of the intervals can be taken into consideration.
If the sign of the interval in their first derivative form gives more than \[0\] then the function is said to be increasing in nature, while if the sign of the intervals in their first derivative form gives less than \[0\] then the function is said to be decreasing in nature.
Finally, we get increasing as well as decreasing intervals of the function.

We are given \[f(x) = {x^2} - x + 1\]
Differentiating both sides with respect to \[x\] we get ,
\[f'(x) = 2x - 1\]
Now, we know that \[f’(x)\] shows the slope and the slope is zero at the maximum and the minimum value points.
Hence, putting \[f'(x) = 0\] ,
Therefore we get \[2x - 1 = 0\]
Hence , \[x = \dfrac{1}{2}\]
The point \[x = \dfrac{1}{2}\] divides the interval \[\left[ {0,1} \right]\]into two disjoint intervals \[\left[ {0,\dfrac{1}{2}} \right]\] and \[\left[ {\dfrac{1}{2},1} \right]\]
Now, substituting any value from the interval \[\left[ {0,\dfrac{1}{2}} \right]\] in the equation of \[f’(x)\] we get a value less than zero. Which signifies that the function is decreasing.
Similarly, substituting any value from the interval \[\left[ {\dfrac{1}{2},1} \right]\] in the equation of \[f’(x)\] we get a value greater than zero. Which signifies that the function is increasing.
Hence, the function decreases for half interval and increases for half interval.
Therefore \[f(x)\] is neither increasing nor decreasing in the interval \[\left[ {0,1} \right]\] .
Therefore option (C) is the correct answer.

Note:
We must know the power rule of differentiation $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$ and the derivatives of some basic trigonometric and algebraic functions for finding the first derivative of the function. We should know that in the intervals in which the derivative is positive, the nature of function is increasing and in the intervals in which the derivative is negative, the function is decreasing. We can find the points where the slope of the graph is zero by equating the derivative to zero and finding the value of the variable.