Question

How do you evaluate the function $p(x) = \dfrac{1}{2}{x^3} + \dfrac{2}{3}{x^2} - \dfrac{1}{4}x + \dfrac{1}{3}$ for $p( - 2)$?

Answer 1

Hint: The given function is $p(x) = \dfrac{1}{2}{x^3} + \dfrac{2}{3}{x^2} - \dfrac{1}{4}x + \dfrac{1}{3}$
Equations written using function notation can also be evaluated. With function notation, you might see a problem like this: $p(x) = \dfrac{1}{2}{x^3} + \dfrac{2}{3}{x^2} - \dfrac{1}{4}x + \dfrac{1}{3}$
You can simply apply what you have already known about evaluating expressions to evaluate a function.
We find the $p( - 2)$.
We just use substitution, addition and division.

Complete step-by-step solution:
The given function is $p(x) = \dfrac{1}{2}{x^3} + \dfrac{2}{3}{x^2} - \dfrac{1}{4}x + \dfrac{1}{3}$
We find the $p( - 2)$
Just put the $x = - 2$on the polynomial equation and you will simplify and you will get the answer.
Let us consider $x = - 2$
The $x$ substitute in the given function, hence we get
\[ \Rightarrow p( - 2) = \dfrac{1}{2}{( - 2)^3} + \dfrac{2}{3}{( - 2)^2} - \dfrac{1}{4}( - 2) + \dfrac{1}{3}\]
We find the square and cube value
First we multiply by $( - 2)$ in three times, hence we get
$ \Rightarrow {( - 2)^3} = - 2 \times - 2 \times - 2 = - 8$
Next we multiply by $( - 2)$ in two times, hence we get
$ \Rightarrow {( - 2)^2} = - 2 \times - 2 = 4$
Now we apply in the function
$ \Rightarrow p( - 2) = \dfrac{1}{2}( - 8) + \dfrac{2}{3}(4) - \dfrac{1}{4}( - 2) + \dfrac{1}{3}$
The First term we multiply $( - 8)$by$1$, the second term we multiply $4$ by $2$, the third term we multiply $( - 2)$ by $( - 1)$, hence we get
$ \Rightarrow p( - 2) = - \dfrac{8}{2} + \dfrac{8}{3} + \dfrac{2}{4} + \dfrac{1}{3}$
Now factor the common term, hence we get
$ \Rightarrow p( - 2) = - \dfrac{8}{2} + \left( {\dfrac{8}{3} + \dfrac{1}{3}} \right) + \dfrac{2}{4}$
We add the numerator in the bracket, hence we get
$ \Rightarrow p( - 2) = - \dfrac{8}{2} + \dfrac{9}{3} + \dfrac{2}{4}$
Divide $9$ by $3$ and divide $8$ by $2$ hence we get
$ \Rightarrow p( - 2) = - 4 + 3 + \dfrac{2}{4}$
Divide the last term
$ \Rightarrow p( - 2) = - 4 + 3 + \dfrac{1}{2}$
Subtract the first term and the second term, hence we get
$ \Rightarrow p( - 2) = - 1 + \dfrac{1}{2}$
Now we take LCM, hence we get
$ \Rightarrow p( - 2) = - \dfrac{2}{2} + \dfrac{1}{2}$
The common term denominator take over the common, hence we get
$ \Rightarrow p( - 2) = \dfrac{{ - 2 + 1}}{2}$
Subtract the numerator, hence we get
$ \Rightarrow p( - 2) = - \dfrac{1}{2}$

Hence the result is $p( - 2) = - \dfrac{1}{2}$.

Note: Evaluating Function notation:
Some people think of functions as ”mathematical machines.” Imagine you have a machine that changes a number according to a specific rule, such as “multiply by$3$and add$2$” or divide by$5$, add$25$, and multiply by$ - 1$.
” If you put a number into the machine, a new number will pop out the other end, having been changed according to the rule.
The number that goes in is called the input, and the number that is produced is called the output. You can also call the machine “$f$” for function.
If you put $x$ into the box, $f(x)$, comes out. Mathematically speaking, $x$ is the input, or the ”independent variable,” and $f(x)$is the output, or “ dependent variable,” since it depends on the value of $x$.