Q. 1 TB - Calculus

Question

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{function} \mathrm{f}\left(\mathrm{x}\right) = {\mathrm{x}}^3-4{\mathrm{x}}^2-12\mathrm{x}. \\ \mathrm{Find} \mathrm{the} \mathrm{intervals} \mathrm{on} \mathrm{which} \mathrm{f} \mathrm{is} \mathrm{positive} \mathrm{and} \mathrm{the} \mathrm{intervals} \mathrm{on} \mathrm{which} \mathrm{f} \mathrm{is} \mathrm{negative}. \end{gathered}$$

Step-by-Step Solution

Answer

The function is positive on interval $\left(-2,0\right)\cup \left(6,\infty \right)$and negative on interval$\left(-\infty ,-2\right)\cup \left(0,6\right)$.

Step 1. Given information

The given function is:

$\mathrm{f}\left(\mathrm{x}\right) = {\mathrm{x}}^3-4{\mathrm{x}}^2-12\mathrm{x}$

Step 2. Finding Zeros of the function

Equate the function to zero and solve the equation for x.

$$\begin{gathered} f\left(x\right)=0 \\ {\mathrm{x}}^3-4{\mathrm{x}}^2-12\mathrm{x} = 0 \\ \mathrm{x}({\mathrm{x}}^2-4\mathrm{x}-12)=0 \\ \mathrm{x} [{\mathrm{x}}^2-6\mathrm{x}+2\mathrm{x}-12]=0 \\ \mathrm{x} \left[\mathrm{x}\right(\mathrm{x}-6)+2(\mathrm{x}-6\left)\right]=0 \\ \mathrm{x} \left[\right(\mathrm{x}-6\left)\right(\mathrm{x}=2\left)\right]=0 \\ \mathrm{x}(\mathrm{x}-6)(\mathrm{x}+2)=0 \\ \Rightarrow \mathrm{x}=0 \mathrm{or} \mathrm{x}=6 \mathrm{or} \mathrm{x}=-2 \end{gathered}$$

So zeros of the function occur at$\mathrm{x}=0, 6, 2$.

Step 3. Checking intervals.

Form interval from zeros of the function.

So intervals are$\left(-\infty ,-2\right), (-2,0), (0,6), (6,\infty )$.

Now take $x=-3$from the interval $\left(-\infty ,-2\right)$ and find the function value.

 $$\begin{gathered} \mathrm{f}(-3)={(-3)}^3-4{(-3)}^2-12(-3) \\ \mathrm{f}(-3)=-27 \end{gathered}$$

Hence the function is negative in the interval$\left(-\infty ,-2\right)$.

Step 3. Checking interval ( - 2 , 0 )

Now take $x=-1$from the interval $(-2,0)$and find the function value.

 $$\begin{gathered} \mathrm{f}(-1)={(-1)}^3-4{(-1)}^2-12(-1) \\ \mathrm{f}(-1)=7 \end{gathered}$$

Hence the function is positive in the interval$(-2,0)$.

Step 3. Checking interval ( 0 , 6 )

Now take $x=1$from the interval $(0,6)$and find the function value.

 $$\begin{gathered} \mathrm{f}\left(1\right)={\left(1\right)}^3-4{\left(1\right)}^2-12\left(1\right) \\ \mathrm{f}\left(1\right)=-15 \end{gathered}$$

Hence the function is negative in the interval$(0,6)$.

Step 3. Checking interval ( 6 , ∞ )

Now take $x=7$from the interval $(6,\infty )$and find the function value.

 $$\begin{gathered} \mathrm{f}\left(7\right)={\left(7\right)}^3-4{\left(7\right)}^2-12\left(7\right) \\ \mathrm{f}\left(7\right)=63 \end{gathered}$$

Hence the function is positive in the interval$(6,\infty )$.