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Question
the function $f(x)=2x^{3}-18x^{2}+30x - 5$ has two critical values. the smaller one equals and the larger one equals
Step1: Find the derivative of f(x)
$f(x)=2x^{3}-18x^{2}+30x - 5$, so $f^\prime(x)=6x^{2}-36x + 30$
Step2: Set the derivative equal to zero
$6x^{2}-36x + 30 = 0$. Divide through by 6: $x^{2}-6x + 5=0$
Step3: Factor the quadratic equation
$(x - 1)(x - 5)=0$
Step4: Solve for x
$x-1=0$ gives $x = 1$; $x - 5=0$ gives $x = 5$
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The smaller one equals 1 and the larger one equals 5.