Browse Questions

# Prove that the function $f$ given by $f (x) = x^2 – x + 1$ is neither strictly increasing nor strictly decreasing on $(– 1, 1).$

Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Let $f(x)=x^2-x+1$
Differentiating w.r.t $x$ we get,
$f'(x)=2x-1$
$\quad=2(x-\large\frac{1}{2})$
Clearly $-1 < x < \large\frac{1}{2}$
$\Rightarrow f'(x) <0$
Step 2:
$\large\frac{1}{2}$$< x < 1\Rightarrow x-\large\frac{1}{2}$$ > 0$
$\Rightarrow 2(x-\large\frac{1}{2})$$> 0\Rightarrow f'(x) > 0$
Thus $f'(x)$ does not have the same sign through out the interval $(-1,1)$
Hence it is neither increasing nor decreasing on $(-1,1)$