# Which of following statements are true or false?Give reasons for your answers.(i)The function f, defined by f(x)=x^3-6x^2+16x-15,is increasing in RR.

Feb 16, 2018

$\setminus$

$\text{See proof below.}$

#### Explanation:

$\setminus$

$\text{To answer this, we will look at} \setminus \setminus f ' \left(x\right) .$

$\text{Given:} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus f \left(x\right) \setminus = \setminus {x}^{3} - 6 {x}^{2} + 16 x - 15.$

$\text{Differentiate:} \setminus q \quad \setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 3 {x}^{2} - 12 x + 16.$

$\text{We want to know whether or not" \ f'(x) \ \ "is always positive.}$

$\text{There are several ways to continue here. Because the function}$
$\text{is a quadratic polynomial with simple coefficients, it might be}$
$\text{easiest just to complete the square here:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 3 {x}^{2} - 12 x + 16$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus 3 \left({x}^{2} - 4 x\right) + 16$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus 3 \left({x}^{2} - 4 x + 4 - 4\right) + 16$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus 3 \left({\left(x - 2\right)}^{2} - 4\right) + 16$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus 3 {\left(x - 2\right)}^{2} - 12 + 16$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus 3 {\left(x - 2\right)}^{2} + 4.$

$\text{So:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 3 {\left(x - 2\right)}^{2} + 4.$

$\text{The expression on the RHS of the previous is clearly positive}$
$\text{everywhere. Here's a proof, if desired:}$

$\setminus q \quad x \setminus \in \mathbb{R} \setminus \quad \implies \setminus \quad$

${\left(x - 2\right)}^{2} \ge 0 , \text{as square of any quantity is always non-negative;}$

$\setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad 3 \setminus \cdot {\left(x - 2\right)}^{2} \ge 3 \setminus \cdot 0 , \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{as 3 is positive;}$

$\setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus 3 {\left(x - 2\right)}^{2} \ge 0 ,$

$\setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus 3 {\left(x - 2\right)}^{2} + 4 \ge 0 + 4 ,$

$\setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus 3 {\left(x - 2\right)}^{2} + 4 \ge 4 ,$

$\setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus 3 {\left(x - 2\right)}^{2} + 4 > 0.$

$\text{Thus:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad x \setminus \in \mathbb{R} \setminus \quad \implies \setminus \quad \setminus 3 {\left(x - 2\right)}^{2} + 4 > 0.$

$\setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus 3 {\left(x - 2\right)}^{2} + 4 \setminus \quad \setminus \text{is positive everywhere.}$

$\setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f ' \left(x\right) \setminus \quad \setminus \text{is positive everywhere.}$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus \quad f \left(x\right) \setminus q \quad \setminus \text{is increasing everywhere.}$

$\setminus$

$\text{I.e.:" \qquad \qquad \qquad \qquad \quad f(x) \ \ "is increasing on" \ \ RR.} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \square$

Feb 16, 2018

The statement is true

#### Explanation:

The first derivative is given by

$f ' \left(x\right) = 3 {x}^{2} - 12 x + 16$

Critical points will occur when $f ' \left(x\right) = 0$.

$0 = 3 {x}^{2} - 12 x + 16$

By the discriminant we get that

${b}^{2} - 4 a c = {\left(- 12\right)}^{2} - 4 \left(3\right) \left(16\right) = - 48$

Since this is negative there will be no solution to the equation. Thus, the function will either be increasing or decreasing for all $x$.

If you test a point, like $x = 0$ into the derivative, you see it will be positive. Thus the function is increasing on all $x$. Therefore, the statement is true.

Hopefully this helps!