Question

If $f\left(x\right)=x^3-10x^2+200x-10$ then

• A. $f\left(x\right)$ is decreasing in $(-\infty ,-10]$ and increasing in $[10,\infty )$
• B. $f\left(x\right)$ is increasing in $(-\infty ,-10]$ and decreasing in $[10,\infty )$
• C. $f\left(x\right)$ is increasing through the number line
• D. $f\left(x\right)$ is decreasing through the number line

Answer 1

The correct option is C

$f\left(x\right)$ is increasing through the number line

Explanation for the correct option:

Find the nature of the function:

For the function $f\left(x\right)=x^3-10x^2+200x-10$, the derivative is given as:

$\begin{array}{rcl}f\text{'}\left(x\right)& =& 3x^2-10\times 2x+200\\ & =& 3x^2-20x+200\\ & =& 3\left(x^2-\frac{20}{3}x+\frac{200}{3}\right)\\ & =& 3\left(x^2-2\times x\times \frac{10}{3}+\frac{100}{9}+\frac{200}{3}-\frac{100}{9}\right)\\ & =& 3\left({\left(x-\frac{10}{3}\right)}^2+\frac{500}{9}\right)\end{array}$

Now as square of a number is always non-negative, so $\begin{array}{l}{\left(x-\frac{10}{3}\right)}^2\end{array}\ge 0$ and thus $3\left(\begin{array}{l}{\left(x-\frac{10}{3}\right)}^2\end{array}+\frac{500}{9}\right)\ge 0$.

Thus for all $x$, it is found that $f\text{'}\left(x\right)>0$. And so the function $f\left(x\right)$ is increasing through the number line.

Hence, the correct option is C.