Calculus I - Functions (Practice Problems)

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Section 1.1 : Functions

For problems 1 – 4 the given functions perform the indicated function evaluations.

f(x)=3−5x−2x2 Solution
1. $f\left( 4 \right)$
2. $f\left( 0 \right)$
3. $f\left( { - 3} \right)$
1. $f\left( {6 - t} \right)$
2. $f\left( {7 - 4x} \right)$
3. $f\left( {x + h} \right)$
g(t)=t2t+6 Solution
1. $g\left( 0 \right)$
2. $g\left( { - 3} \right)$
3. $g\left( {10} \right)$
1. $g\left( {{x^2}} \right)$
2. $g\left( {t + h} \right)$
3. $g\left( {{t^2} - 3t + 1} \right)$
h(z)=√1−z2 Solution
1. $h\left( 0 \right)$
2. $h\left( { - \frac{1}{2}} \right)$
3. $h\left( {\frac{1}{2}} \right)$
1. $h\left( {9z} \right)$
2. $h\left( {{z^2} - 2z} \right)$
3. $h\left( {z + k} \right)$
R(x)=√3+x−4x+1 Solution
1. $R\left( 0 \right)$
2. $R\left( 6 \right)$
3. $R\left( { - 9} \right)$
1. $R\left( {x + 1} \right)$
2. $R\left( {{x^4} - 3} \right)$
3. $R\left( {\frac{1}{x} - 1} \right)$

The difference quotient of a function $f\left( x \right)$ is defined to be,

$\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

For problems 5 – 9 compute the difference quotient of the given function.

$f\left( x \right) = 4x - 9$ Solution
$g\left( x \right) = 6 - {x^2}$ Solution
$f\left( t \right) = 2{t^2} - 3t + 9$ Solution
$\displaystyle y\left( z \right) = \frac{1}{{z + 2}}$ Solution
$\displaystyle A\left( t \right) = \frac{{2t}}{{3 - t}}$ Solution

For problems 10 – 17 determine all the roots of the given function.

$f\left( x \right) = {x^5} - 4{x^4} - 32{x^3}$ Solution
$R\left( y \right) = 12{y^2} + 11y - 5$ Solution
$h\left( t \right) = 18 - 3t - 2{t^2}$ Solution
$g\left( x \right) = {x^3} + 7{x^2} - x$ Solution
$W\left( x \right) = {x^4} + 6{x^2} - 27$ Solution
$f\left( t \right) = {t^{\frac{5}{3}}} - 7{t^{\frac{4}{3}}} - 8t$ Solution
$\displaystyle h\left( z \right) = \frac{z}{{z - 5}} - \frac{4}{{z - 8}}$ Solution
$\displaystyle g\left( w \right) = \frac{{2w}}{{w + 1}} + \frac{{w - 4}}{{2w - 3}}$ Solution

For problems 18 – 22 find the domain and range of the given function.

$Y\left( t \right) = 3{t^2} - 2t + 1$ Solution
$g\left( z \right) = - {z^2} - 4z + 7$ Solution
$f\left( z \right) = 2 + \sqrt {{z^2} + 1}$ Solution
$h\left( y \right) = - 3\sqrt {14 + 3y}$ Solution
$M\left( x \right) = 5 - \left| {x + 8} \right|$ Solution

For problems 23 – 32 find the domain of the given function.

$\displaystyle f\left( w \right) = \frac{{{w^3} - 3w + 1}}{{12w - 7}}$ Solution
$\displaystyle R\left( z \right) = \frac{5}{{{z^3} + 10{z^2} + 9z}}$ Solution
$\displaystyle g\left( t \right) = \frac{{6t - {t^3}}}{{7 - t - 4{t^2}}}$ Solution
$g\left( x \right) = \sqrt {25 - {x^2}}$ Solution
$h\left( x \right) = \sqrt {{x^4} - {x^3} - 20{x^2}}$ Solution
$\displaystyle P\left( t \right) = \frac{{5t + 1}}{{\sqrt {{t^3} - {t^2} - 8t} }}$ Solution
$f\left( z \right) = \sqrt {z - 1} + \sqrt {z + 6}$ Solution
$\displaystyle h\left( y \right) = \sqrt {2y + 9} - \frac{1}{{\sqrt {2 - y} }}$ Solution
$\displaystyle A\left( x \right) = \frac{4}{{x - 9}} - \sqrt {{x^2} - 36}$ Solution
$Q\left( y \right) = \sqrt {{y^2} + 1} - \sqrt[3]{{1 - y}}$ Solution

For problems 33 – 36 compute $\left( {f \circ g} \right)\left( x \right)$ and $\left( {g \circ f} \right)\left( x \right)$ for each of the given pair of functions.

$f\left( x \right) = 4x - 1$ , $g\left( x \right) = \sqrt {6 + 7x}$ Solution
$f\left( x \right) = 5x + 2$ , $g\left( x \right) = {x^2} - 14x$ Solution
$f\left( x \right) = {x^2} - 2x + 1$ , $g\left( x \right) = 8 - 3{x^2}$ Solution
$f\left( x \right) = {x^2} + 3$ , $g\left( x \right) = \sqrt {5 + {x^2}}$ Solution