Q. 3 - Calculus

Question

Sign analyses: For each of the following functions g(x), use algebra and a sign chart to find the intervals on which g(x) is positive and the intervals on which g(x) is negative.

$$\begin{gathered} \left(a\right) g\left(x\right)=6x^2-18x \\ \left(b\right) g\left(x\right)=2-x-2x^2+x^3 \\ \left(c\right) g\left(x\right)=\frac{(x+2)(x-1{)}^4}{e^x} \\ \left(d\right) g\left(x\right)=\frac{3x^2-5x-2}{{\sin }^{2}x} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The given function is positive in the intervals $(-\mathrm{\infty },0)\text{ and }(3,\mathrm{\infty }),$ and negative in the interval $(0,3).$

Part (b) The given function is positive in the intervals $(-1,1)\text{ and }(2,\mathrm{\infty }),$ and negative in the intervals $(-\mathrm{\infty },-1)\text{ and }(1,2).$

Part (c) The given function is positive in the interval $(2,\mathrm{\infty }),$ and negative in the interval $(-\mathrm{\infty },-2).$

Part (d) The given function is positive in the intervals $(-\mathrm{\infty },1)\text{ and }(\frac{2}{3},\mathrm{\infty }),$ and negative in the interval $\left(1,\frac{2}{3}\right).$

Part (a) Step 1. Given Information.

The given function is $g\left(x\right)=6x^2-18x.$

Part (a) Step 2. Find the intervals.

The given function is positive for the values of x when g(x) > 0, and negative for the values of x when g(x) < 0.

Now, to find the intervals put g(x) = 0, 

$$\begin{gathered} g\left(x\right)=6x^2-18x \\ 0=6x\left(x-3\right) \\ x=0 \mathrm{and} x=3 \end{gathered}$$

Sign chart of the function is 

Interval	x	(x - 3)	g(x)
x > 3	+	+	+
0	+	-	-
x < 0	-	-	+

Thus, the function is positive in the intervals $(-\mathrm{\infty },0)\text{ and }(3,\mathrm{\infty }),$ and negative in the intervals $\left(0,3\right).$

Part (b) Step 1. Find the intervals.

The given function is positive for the values of x when g(x) > 0, and negative for the values of x when g(x) < 0.

Now, to find the intervals put g(x) = 0, 

$$\begin{gathered} g\left(x\right)=2-x-2x^2+x^3 \\ 0=2-x-2x^2+x^3 \\ 0=\left(x-1\right)\left(x+1\right)\left(x-2\right) \\ x=1, x=-1, \mathrm{and} x=2 \end{gathered}$$

Sign chart of the function is 

Interval	(x-2)	(x - 1)	(x + 1)	g(x)
x < -1	-	-	-	-
-1	-	-	+	+
1	-	+	+	-
x > 2	+	+	+	+

Thus, the function is positive in the intervals $(-1,1)\text{ and }(2,\mathrm{\infty }),$ and negative in the intervals $(-\mathrm{\infty },-1)\text{ and }(1,2).$

Part (c) Step 1. Find the intervals.

The given function is positive for the values of x when g(x) > 0, and negative for the values of x when g(x) < 0.

Since the function $e^x\text{ and }(x-1{)}^4$ are always on the real line, thus g(x) > 0,

$$\begin{gathered} \left(x+2\right)>0 \\ x>-2 \end{gathered}$$

And g(x) < 0,

$$\begin{gathered} \left(x+2\right)<0 \\ x<-2 \end{gathered}$$

Sign chart of the function is 

Interval	(x-2)	g(x)
x < -2	-	-
-2 < x < 1	-	-
1	-	-
x > 2	+	+

Thus, the function is positive in the interval $(2,\mathrm{\infty }),$ and negative in the interval $(-\mathrm{\infty },-2).$

Part (d) Step 1. Find the intervals.

The given function is positive for the values of x when g(x) > 0, and negative for the values of x when g(x) < 0.

Since the function ${\sin }^{2}x$ are always on the real line, g(x) > 0 and g(x) < 0,

$$\begin{gathered} 3x^2-5x-2>0 \\ \left(x-1\right)\left(3x-2\right)>0 \\ \mathrm{And} \\ 3{\mathrm{x}}^2-5\mathrm{x}-2<0 \\ \left(\mathrm{x}-1\right)\left(3\mathrm{x}-2\right)<0 \end{gathered}$$

To find the intervals let's put g(x) = 0,

$$\begin{gathered} \left(x-1\right)\left(3x-2\right)=0 \\ x=1 \mathrm{and} x=\frac{2}{3} \end{gathered}$$

Sign chart of the function is 

Interval	(x-1)	$\left(x-\frac{2}{3}\right)$	$3x^2-5x-2$	g(x)
x < 1	-	-	+	+
$1<x<\frac{2}{3}$	+	-	-	-
$\frac{2}{3}<x<8$	+	+	+	+

Thus, the function is positive in the intervals $(-\mathrm{\infty },1)\text{ and }(\frac{2}{3},\mathrm{\infty })$ and negative in the interval $\left(1,\frac{2}{3}\right).$