\
"To answer this, we will look at" \ \ f'(x).
"Given:" \qquad \qquad \qquad \qquad \qquad \quad \ f(x) \ = \ x^3 - 6 x^2 + 16 x - 15.
"Differentiate:" \qquad \qquad \quad f'(x) \ = \ 3 x^2 - 12 x + 16.
"We want to know whether or not" \ f'(x) \ \ "is always positive."
"There are several ways to continue here. Because the function"
"is a quadratic polynomial with simple coefficients, it might be"
"easiest just to complete the square here:"
\qquad \qquad \qquad \qquad \qquad \qquad \quad f'(x) \ = \ 3 x^2 - 12 x + 16
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 3 ( x^2 - 4 x ) + 16
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 3 ( x^2 - 4 x + 4 - 4 ) + 16
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 3 ( ( x - 2 )^2 - 4 ) + 16
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 3 ( x - 2 )^2 - 12 + 16
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 3 ( x - 2 )^2 + 4.
"So:"
\qquad \qquad \qquad \qquad \qquad \qquad \quad f'(x) \ = \ 3 ( x - 2 )^2 + 4.
"The expression on the RHS of the previous is clearly positive"
"everywhere. Here's a proof, if desired:"
\qquad x \in RR \quad => \quad
( x - 2 )^2 >= 0, "as square of any quantity is always non-negative;"
\qquad \qquad \qquad :. \qquad \qquad \quad \quad 3 \cdot ( x - 2 )^2 >= 3 \cdot 0, \qquad \qquad \qquad \quad "as 3 is positive;"
\qquad \qquad \qquad :. \qquad \qquad \qquad \quad \ 3 ( x - 2 )^2 >= 0,
\qquad \qquad \qquad :. \qquad \qquad \qquad \quad \ 3 ( x - 2 )^2 + 4 >= 0 + 4,
\qquad \qquad \qquad :. \qquad \qquad \qquad \quad \ 3 ( x - 2 )^2 + 4 >= 4,
\qquad \qquad \qquad :. \qquad \qquad \qquad \quad \ 3 ( x - 2 )^2 + 4 > 0.
"Thus:"
\qquad \qquad \qquad \qquad \qquad \qquad x \in RR \quad => \quad \ 3 ( x - 2 )^2 + 4 > 0.
\qquad :. \qquad \qquad \ 3 ( x - 2 )^2 + 4 \quad \ "is positive everywhere."
\qquad :. \qquad \qquad \qquad \qquad f'(x) \quad \ "is positive everywhere."
\qquad \qquad :. \qquad \qquad \quad f(x) \qquad \ "is increasing everywhere."
\
"I.e.:" \qquad \qquad \qquad \qquad \quad f(x) \ \ "is increasing on" \ \ RR." \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ square
