Question

question 17 (5 points)
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for which interval(s) is the function increasing and decreasing?
$y = 3x^{3}-16x + 2$
increasing for $-2.31lt xlt2.31$; decreasing for $xlt - 2.31$ and $xgt2.31$
increasing for $xlt-\frac{4}{3}$ and $xgt\frac{4}{3}$; decreasing for $-\frac{4}{3}lt xlt\frac{4}{3}$
increasing for $xlt - 2.31$ and $xgt2.31$; decreasing for $-2.31lt xlt2.31$
increasing for $-\frac{4}{3}lt xlt\frac{4}{3}$; decreasing for $xlt-\frac{4}{3}$ and $xgt\frac{4}{3}$

Explanation:

Step1: Find the derivative

The derivative of \(y = 3x^{3}-16x + 2\) is \(y'=9x^{2}-16\).

Step2: Set the derivative equal to zero

Set \(y'=0\), so \(9x^{2}-16 = 0\). Then \(9x^{2}=16\), \(x^{2}=\frac{16}{9}\), \(x=\pm\frac{4}{3}\approx\pm1.33\) (not \(\pm2.31\)).

Step3: Analyze the sign of the derivative

For \(y'=9x^{2}-16=(3x - 4)(3x + 4)\):

• When \(x<-\frac{4}{3}\), let \(x=-2\), then \(y'=9\times(-2)^{2}-16=36 - 16=20>0\).
• When \(-\frac{4}{3}
• When \(x>\frac{4}{3}\), let \(x = 2\), then \(y'=9\times2^{2}-16=36 - 16=20>0\).

Answer:

increasing for \(x<-\frac{4}{3}\) and \(x>\frac{4}{3}\); decreasing for \(-\frac{4}{3}