Evaluating Functions

A function is a mapping between an input and an output. For example, the function $f(x) = x^2$ takes an input $x$ and returns its square $x^2.$ In this case, $f(2) = 2^2 = 4,$ $f\big(\sqrt{3}\big) = \big(\sqrt{3}\big)^2 = 3,$ and so on. The key difference between a function and a more general relation is that for every input to a function, there is exactly one output.

What is the value of the function $f(x) = 3x - 5$ at $x = 5$?

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We see that we want to evaluate $f(5)$, where $f(x) = 3x-5$. Thus,

$f(5) = 3(5)-5 = 15-5 = 10. \ _\square$

Sometimes when mapping between an input and output, the input can be another function that maps to another input. This is called a composite function. When evaluating a composite function, first we compose the function and evaluate the result as we do any other function.

Given that $f(x)=\frac{3x}{x-1}$ and $g(x)=\frac{x}{1-2x}$ what is the value of the composite function $(f \circ g)(x)$ at $x=1$?

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First compose the function:

$\begin{aligned} (f \circ g)(x)= f\big(g(x)\big)=f\left( \frac { x }{ 1-2x } \right) &=\frac { 3\left( \frac { x }{ 1-2x } \right) }{ \frac { x }{ 1-2x } -1 }\\\\ &=\frac { 3x }{ 3x-1 }. \end{aligned}$

Now, evaluate the function $f\big(g(x)\big)=\frac { 3x }{ 3x-1}$ at $x=1:$

$f\big(g(1)\big)=\frac{3(1)}{3(1)-1}=\frac{3}{2}.\ _\square$

Given that $f(x)=x^{2}-1, g(x) = x+1,$ and $h(x)=2x$, what is the value of the function $(f\circ g\circ h)(x)$ at $x=3$?

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First compose the function:

$\begin{aligned} (f\circ g\circ h)(x)=f\Big(g\big(h(x)\big)\Big) &=f\big(h(x)+1\big)\\ &={ \big(h(x)+1\big) }^{ 2 }-1\\ &={ (2x+1) }^{ 2 }-1\\ &=4{ x }^{ 2 }+4x. \end{aligned}$

Now, evaluate $f\Big(g\big(h(x)\big)\Big)=4x^{2}+4x$ at $x=3:$

$4(3)^{2}+4(3)=48.\ _\square$

Given that $f(x)=ax+3$ and $g(x)=3x$, for what value of $a$ is $(f\circ g)(4)=9$?

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Compose the function and evaluate at $4:$

$(f\circ g)(4)=f\big(g(4)\big)=12a+3.$

Equate this to $9$ to obtain $12a+3=9 \implies a=\frac { 1 }{ 2 } .\ _\square$

What is the value of the function $f(x)=(x-3)(x+13)(x-4)(x-6)+23$ at $x=4$?

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We could just plug in $4$ in every place of $x,$ but notice that $x-4=0$ when $x=4,$ which will collapse all the product terms with it. Thus,

$(x-3)(x+13)(0)(x-6)+23=0+23=23.\ _\square$

Sometimes a function is given as a piecewise defined function, which is a function defined by multiple sub-functions. Each sub-function is defined by a certain interval or conditions.

If function $f$ is defined by $f(x) = \begin{cases} 3x - 2 & , x > 3 \\ x^2 - 2 & , -2 \leq x \leq 2 \\ 2x + 1 & , x < -3 \\ \end{cases}$ then find the values, if exists, of

$\begin{aligned} (i) f(4) & & (ii) f(2.5) \\ (iii) f(-2) & & (iv) f(-4) \\ (v) f(0) & & (vi) f(-7) \\ \end{aligned}$

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Note that the domain of $f$ is $(- \infty , -3) \cup [-2,2] \cup (3, \infty)$

$(i)$ Since $f(x) = 3x - 2$ for $x > 3$, we have $f(4) = 3 \cdot 4 - 2 = 12 - 2 = 10$

$(ii)$ 2.5 does not belong to domain $f$, $f(2.5)$ is not defined.

$(iii)$ Since $f(x) = x^2 - 2, -2 \leq x \leq 2$, we have $f(-2) = (-2)^2 - 2 = 4 - 2 = 2$

$(iv)$ Since $f(x) = 2x + 1, x \leq -3$, we have $f(-4) = 2(-4) + 1 = -7$

$(v)$ Since $f(x) = x^2 - 2, -2 \leq x \leq 2$, we have $f(0) = (0)^2 - 2 = -2$

$(vi)$ Since $f(x) = 2x + 1, x \leq -3$, we have $f(-7) = 2(-7) + 1 = -14 + 1 = -13$

Given the following function: $f(x)= \begin{cases} 3x+1 & x>3 \\ 2x & -1\le x\le 3\\ 3 & x<-1, \end{cases}$ what is the value of $f(7)+f(3)+f(0)+f(-100)$?

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Based on the piecewised function above, if $x>3,$ we evaluate over the function $f(x)=3x+1.$ If $x$ is between $-1$ and $3$ inclusive, then we evaluate over the function $f(x)=2x.$ If $x<-1,$ we evaluate over the function $f(x)=3.$

$\begin{aligned} f( 7) +f(3) +f( 0 ) +f( -100 ) &=\big(3(7)+1\big)+\big(2(3)\big)+(2(0))+3\\ &=22+6+3\\ &=31.\ _\square \end{aligned}$