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