Question

Find the value of $k\left(0\right),k\left(3\right),k\left(-3\right),k\left(\frac{1}{2}\right),k\left(\frac{a}{2}\right),k\left(-x\right),k\left(x^3\right)$for the function $k\left(x\right)=2x^3-3x^2$.

Short answer

The value of function are

$\begin{array}{l}k\left(0\right)=0,k\left(3\right)=27,k\left(-3\right)=-81,k\left(\frac{1}{2}\right)=-\frac{1}{2},k\left(\frac{a}{2}\right)=\frac{a^3}{4}-\frac{3a^2}{4},k\left(-x\right)=-2x^3-3x^2,\\ k\left(x^3\right)=2x^9-3x^6\end{array}$.

Step-by-step solution

Step 1. Evaluate k0,k3,k−3,k12 for the given function.

From the given function,

$k\left(x\right)=2x^3-3x^2\mathrm{....}\left(1\right)$

Evaluate the value of function at $x=0,3,-3,\frac{1}{2}$gives,

role="math" localid="1644515048784" $$\begin{gathered} k\left(0\right)=2{\left(0\right)}^3-3{\left(0\right)}^2\mathrm{....}\left(\text{Substitute}x=0\text{in}\left(1\right)\right) \\ =0 \\ k\left(3\right)=2{\left(3\right)}^3-3{\left(3\right)}^2\mathrm{....}\left(\text{Substitute}x=3\text{\hspace{0.17em}in}\left(1\right)\right) \\ =27 \\ k\left(-3\right)=2{\left(-3\right)}^3-3{\left(-3\right)}^2\mathrm{....}\left(\text{Substitute}x=-3\text{in}\left(1\right)\right) \\ =-81 \\ k\left(\frac{1}{2}\right)=2{\left(\frac{1}{2}\right)}^3-3{\left(\frac{1}{2}\right)}^2\mathrm{....}\left(\text{Substitute}x=\frac{1}{2}\text{in}\left(1\right)\right) \\ =-\frac{1}{2} \end{gathered}$$

Step 2. Evaluate ka2,k−x,kx3 for the given function.

Similarly, evaluate the value of function (1) at $x=\frac{a}{2},-x,x^3$gives,

role="math" localid="1644515366009" $$\begin{gathered} k\left(\frac{a}{2}\right)=2{\left(\frac{a}{2}\right)}^3-3{\left(\frac{a}{2}\right)}^2\mathrm{....}\left(\text{Substitute}x=\frac{a}{2}\text{in}\left(1\right)\right) \\ =\frac{a^3}{4}-\frac{3a^2}{4} \\ k\left(-x\right)=2{\left(-x\right)}^3-3{\left(-x\right)}^2\mathrm{....}\left(\text{Substitute}x=-x\text{\hspace{0.17em}in}\left(1\right)\right) \\ =-2x^3-3x^2 \\ k\left(x^3\right)=2{\left(x^3\right)}^3-3{\left(x^3\right)}^2....\left(\text{Substitute}x=x^3\text{in}\left(1\right)\right) \\ =2x^9-3x^6 \end{gathered}$$

Step 3. Description of steps.

From the obtained values,

$\begin{array}{l}k\left(0\right)=0,k\left(3\right)=27,k\left(-3\right)=-81,k\left(\frac{1}{2}\right)=-\frac{1}{2},k\left(\frac{a}{2}\right)=\frac{a^3}{4}-\frac{3a^2}{4},k\left(-x\right)=-2x^3-3x^2,\\ k\left(x^3\right)=2x^9-3x^6\end{array}$.