Problem:
Find the largest and smallest critical numbers of the function
[tex]f(x)=2x^3-21x^2+36x-2.[/tex]
Solution Steps:
1. Find the Derivative:
To find the critical numbers of the function, we first need to
determine [tex]f'(x).[/tex]
Given:
[tex]f(x)=2x^3-21x^2+36x-2[/tex]
Differentiating with respect to [tex]x[/tex] on both sides,
[tex]\dfrac{df(x)}{dx}=\dfrac{d}{dx}(2x^3-21x^2+36x-2)[/tex]
[tex]f'(x)=\dfrac{d}{dx}(2x^3)-\dfrac{d}{dx}(21x^2)+\dfrac{d}{dx}(36x)-\dfrac{d}{dx}(2)\\[/tex]
[tex]f'(x)=6x^2-42x+36[/tex]
2. Set the derivative equal to zero:
To find the critical numbers, set [tex]f'(x)[/tex] to zero and solve for [tex]x[/tex]:
[tex]6x^2-42x+36=0[/tex]
3. Simplify the equation:
Divide the entire equation by 6 to simplify:
[tex]x^2-7x+6=0[/tex]
4. Factor the quadratic equation:
[tex]x^2-7x+6=(x-1)(x-6)=0[/tex]
So the solutions are:
[tex]x=1[/tex]
[tex]x=6[/tex]
These are the critical numbers of the function [tex]f(x)[/tex].
Conclusion:
The smallest critical number of the function [tex]f(x)=2x^3-21x^2+36x-2[/tex] is [tex]x=1[/tex] and the largest critical number is [tex]x=6.[/tex]
