Chapter 4.4, Problem 43E

To determine

To identify whether the given function is ever negative.

Answer to Problem 43E

No, the give function is not ever negative.

Explanation of Solution

Given information : The equation of the function , that is, $f\left(x\right)=x^2-x+1$ is given.

Differentiate the function $f\left(x\right)=x^2-x+1$ with respect to x on both sides.

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(x^2-x+1\right)\\ f^{\prime }\left(x\right)=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(1\right)\\ =2x-1+0\\ =2x-1\end{array}$

Now, equate the obtained derivative to zero to get the critical point of the given function where the function might have minimal or maxima.

  $\begin{array}{c}2x-1=0\\ x=\frac{1}{2}\end{array}$

Determine the second derivative of the given function by differentiating the obtained first derivative with respect to x on both sides.

  $\begin{array}{c}\frac{d}{dx}\left(f^{\prime }\left(x\right)\right)=\frac{d}{dx}\left(2x-1\right)\\ f^{″}\left(x\right)=2\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(1\right)\\ =2\left(1\right)-0\\ =2\end{array}$

Determine the value of the obtained second derivative at $x=\frac{1}{2}$ .

  $\begin{array}{c}{f}^{″}\left(\frac{1}{2}\right)=2\\ >0\end{array}$

Since the second derivative is positive at the obtained critical point $x=\frac{1}{2}$ , the minima will occur at $x=\frac{1}{2}$ .

Determine the value of the function at $x=\frac{1}{2}$ .

  $\begin{array}{c}f\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-\left(\frac{1}{2}\right)+1\\ =\frac{1}{4}-\frac{1}{2}+1\\ =\frac{1-2+4}{4}\\ =\frac{3}{4}\end{array}$

As a result, the minima will occur at $\left(\frac{1}{2},\frac{3}{4}\right)$ .

This implies that,

  $f\left(x\right)\ge \frac{3}{4}>0$

As a result, $f\left(x\right)>0\forall x$

Therefore, the function $f\left(x\right)=x^2-x+1$ which is a quadratic function with positive coefficient of $x^2$ and for which the minima occur at $\left(\frac{1}{2},\frac{3}{4}\right)$ , the function can never be negative.