Question

Show that the function \[f\left( x \right) = {x^3} - 3{\rm{ }}{x^2} + 6x - 100\] is increasing on R.

Answer 1

Hint: In this question you have to show that the function is increasing on R. R is referred to real number. As we all know, real numbers can be defined as the union of both rational and irrational numbers. We will show that the given function increases with increasing R.

Step wise solution:
Given data: The function is given by \[f\left( x \right) = {x^3} - 3{\rm{ }}{x^2} + 6x - 100\]
We need to show that f(x) is increasing on R.
We will differentiate the function f(x) with respect to x.
Here , \[f\left( x \right) = {x^3} - 3{\rm{ }}{x^2} + 6x - 100\]
Differentiating both sides with respect to x, we get
\[
\dfrac{d}{{dx}}(f\left( x \right)) = \dfrac{d}{{dx}}({x^3} - 3{\rm{ }}{x^2} + 6x - 100)\\
 \Rightarrow f'(x) = \dfrac{d}{{dx}}({x^3}) - \dfrac{d}{{dx}}(3{x^2}) + \dfrac{d}{{dx}}(6x) - \dfrac{d}{{dx}}(100)\\
 \Rightarrow f'(x) = 3{x^2} - 3 \times 2x + 6 - 0\\
 \Rightarrow f'(x) = 3{x^2} - 6x + 6 \\ \]
To show that \[f'\left( x \right) > 0\]
We have get ,
\[
 \Rightarrow f'(x) = 3{x^2} - 6x + 6\\
 \Rightarrow f'(x) = 3({x^2} - 2x + 2) \\
\]
Here, you can see \[({x^2} - 2x + 2)\] is a quadratic polynomial in the form of \[A{x^2} + Bx + C\,\,where\,\,A = 1,B = - 2,C = 2\]
To find out zeros of the polynomial \[{x^2} - 2x + 2\] , we will use discriminant \[{b^2} - 4ac = D\]
So,
\[
 = {b^2} - 4ac = {( - 2)^2} - 4 \times 1 \times 2\\
 \Rightarrow D = {b^2} - 4ac = 4 - 8\\
 \Rightarrow D = - 4\\
 \Rightarrow D < 0 \\ \]
I.e.: the zeros of \[{x^2} - 2x + 2\] are complex. And because, discriminant is less than 0 and the coefficient of \[{x^2}\] is greater than 0, it gives \[({x^2} - 2x + 2) > 0\]

i.e.\[:D < 0 \,\,\ gives\,\,{x^2} - 2x + 2 > 0\\
coefficient\,\,{x^2} > 0 \]

Now,
\[f'(x) = 3({x^2} - 2x + 2)\]
Since , \[3 > 0\,\,and\,\,{x^2} - 2x + 2 > 0,f'(x)\] is also greater than zero.
\[
f'(x) = 3({x^2} - 2x + 2) \\
f'(x) > 0 for\,\,all\,\,x \in R \\ \]

To show that f(x) is increasing on R.
You have seen we have get \[ f'(x) > 0 for\,\,all\,\,x \in R \\ \]
Let us consider a range (a,b) where a\[
f(y) - f(x) = f'(c)(y - x) > 0\\
f(y) - f(x) > 0\\ \]
i.e.:\[f(y) > f(x)\,\,\,for\,\,y > x\,\, \]

So, f is strictly increasing on (a,b)
Hence, \[f\left( x \right) = {x^3} - 3{\rm{ }}{x^2} + 6x - 100\] is increasing on R.

Note: In part 2 you can also write \[3({x^2} - 2x + 2)\,\,as\,\,3({(x - 1)^2} + 1)\] , by this way you can show that \[f'(x) > 0\] . Because \[3 > 0,{\rm{ }}{\left( {x - 1} \right)^2} \] is a square number. So \[3({(x - 1)^2} + 1) = f'(x)\] is greater than 0. The real number includes natural numbers or counting numbers.