একটি রৈখিক ও বক্র রৈখিক ফাংশনের নিয়ম মূল্যায়ন করো

Plug It In, Plug It In

Just like with the old Glade commercials, the key to evaluating functions is to "plug it in, plug it in."

(Video) 2004 Glade Plug-In Commercial

by 513official4

Suppose we're given the function: $$f(x)=3x-2$$ And we're asked to evaluate the function when $x=10$. All you have to do is plug in $10$ everywhere you see an $x$. $$\begin{align} f(x)&=3x-2\\ f({\color{#29A603}{10}})&=3({\color{#29A603}{10}})-2\\ f(10)&=30-2\\ f(10)&=28 \end{align}$$

Similarly, you might be asked to find $f(10)$. This is the same thing as asking to evaluate the function when $x=10$. Let's look at another example: $$f(x)=\frac12x+9$$ Evaluate the function at $f(12)$.

Since the $12$ is in parentheses, we know that $x=12$. So, like before, we plug in $12$ everywhere there's an $x$. $$\begin{align} f(x)&=\frac12x+9\\ f({\color{#29A603}{12}})&=\frac12({\color{#29A603}{12}})+9\\ f(12)&=6+9\\ f(12)&=15 \end{align}$$

The final type of question we can run into is one where they tell us what $f(x)$ equals and we need to find $x$. Suppose we have this function: $$f(x)=\frac{3x}{5}$$ Now, we want to find $x$ when $f(x)=12$. To solve these, we just write out the equation and solve for the missing variable, like we've done in the past. $$\begin{align} f(x)&=\frac{3x}5\\ \color{#29A603}{12}&=\frac{3x}{5}\\ 12\,\color{#29A603}{\times\,5}&=\frac{3x}{5}\,\color{#29A603}{\times\,5}\\ 60&=3x\\ \frac{60}{\color{#29A603}{3}}&=\frac{3x}{\color{#29A603}{3}}\\ 20&=x \end{align}$$ So when $f(x)=12$, $x=20$. We can also write this as: $$f(20)=12$$

Since we want to find $h(-2)$, we want to plug $-2$ into all the spots where there's an $x$. $$\begin{align} h(x)&=x-2x+9x\\ h(-2)&={\color{#29A603}{-2}}-2({\color{#29A603}{-2}})+9({\color{#29A603}{-2}})\\ &=-2+4-18\\ h(-2)&=-16 \end{align}$$

Evaluating Functions

Evaluating functions is actually much easier (and less judgmental) than it sounds.

When you evaluate a function, you plug in a numerical value for the variable in the function. You are essentially putting an input into the function and seeing what the output is.

Another Example

Let's work with the function below:

You'll remember from before when we learned about functions, that the left side of the equation, $f(x)$, represents $x$ as a function of $y$.

$$f(x) = 2x^2 -3x + 1\\ \text{where }x = -1$$ $$\begin{align} f(-1) &= 2({\color{#29A603}{-1}})^2 - 3({\color{#29A603}{-1}}) + 1\\ f(-1) &= 2(1) + 3 + 1\\ f(-1) &= 2 + 3 + 1\\ f(-1) &= 6 \end{align}$$

Let's use a linear function for the next example:

$$f(a) = -5a + 7\\ \text{where }a = 2$$ $$\begin{align} f(2) &= -5({\color{#29A603}{2}}) + 7\\ f(2) &= -10 + 7\\ f(2) &= -3 \end{align}$$

$$f(a) = -5a + 7\\ \text{where }a = -6$$ $$\begin{align} f(2) &= -5({\color{#29A603}{-6}}) + 7\\ f(2) &= 30 + 7\\ f(2) &= 37 \end{align}$$

It's important to note, if you have expressions in parentheses, you're going to want to use the distributive property to expand, and then evaluate the expressions! This is the best process for simplifying to find the solution of the equation.

(Video) Linear Functions

by Mahalo

In this video, Allison explains function notation and how to evaluate a function for a given value.

$\large{\text{Summary}}$

Simply take an equation and plug a value in for the variable, then simplify until there's just one number left. Like so:

$$f(x) = 7x - 2\\ \text{where }x = 8$$ $$\begin{align} f(8) &= 7({\color{#29A603}{8}}) - 2\\ f(8) &= 56 - 2\\ f(8) &= 54 \end{align}$$

This sequence means that for the function $f(x) = 7x - 2$, the input $8$ will always result in the output $54$.

The same goes for non-linear functions: each input will have no more than one output. If you are evaluating any function at a given value, simply plug in the value everywhere you see that variable (usually $x$).

$$f(t) = 2t -17\\ \text{where }t = 6$$ $$\begin{align} f(6) &= 2({\color{#29A603}{6}}) -17\\ f(6) &= 12 -17\\ f(6) &= -5 \end{align}$$