Question

closed interval method: problem 1
(1 point)
the function ( f(x)=2 x^{3}-18 x^{2}+30 x - 11 ) has two critical values. the smaller one equals ( square ) and the larger one equals ( square )
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Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=2x^{3}-18x^{2}+30x - 11\) is \(f^{\prime}(x)=6x^{2}-36x + 30\) (using the power rule \((x^{n})^\prime=nx^{n - 1}\)).

Step2: Set the derivative equal to zero

Set \(f^{\prime}(x)=0\), so \(6x^{2}-36x + 30 = 0\). Divide through by \(6\) to get \(x^{2}-6x + 5=0\).

Step3: Factor the quadratic equation

Factor \(x^{2}-6x + 5=(x - 1)(x - 5)=0\).

Step4: Solve for \(x\)

Using the zero - product property \(x-1 = 0\) gives \(x = 1\) and \(x - 5=0\) gives \(x = 5\).

Answer:

The smaller one equals \(1\) and the larger one equals \(5\).