Question

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.

Answer 1

`\`

`"See proof below."`

Explanation:

`\`

`"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`

Answer 2

The statement is true

Explanation:

The first derivative is given by

`f'(x) = 3x^2 - 12x + 16`

Critical points will occur when `f'(x) =0`.

`0 = 3x^2 -12x + 16`

By the discriminant we get that

`b^2 -4ac = (-12)^2 - 4(3)(16) = -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!