Question

Determine the range of the function $\displaystyle f(x)=a\sqrt[3]{bx-c}+d$.​

Answer 1

Basic Concept Used :-

How to find the range of function :

Given :- f(x)

• 1. Put y = f(x)

• 2. Solve the equation y = f(x) to get x in terms of y, Let assume that x = g(y)

• 3. Now, Find the domain of g(y)

• 4. The values thus obtained for y is the Range of the function f(x).

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:f(x) = a \sqrt[3]{bx - c} + d$

To find the range of f(x),

$\rm :\longmapsto\:Let \: f(x) = y = a \sqrt[3]{bx - c} + d$

$\rm :\longmapsto\: y = a \sqrt[3]{bx - c} + d$

$\rm :\longmapsto\: y - d= a \sqrt[3]{bx - c}$

$\rm :\longmapsto\: \dfrac{y - d}{a} \: = \: \sqrt[3]{bx - c}$

On cubing both sides, we get

$\rm :\longmapsto\: {\bigg(\dfrac{y - d}{a} \bigg) }^{3} = bx - c$

$\rm :\longmapsto\: {\bigg(\dfrac{y - d}{a} \bigg) }^{3} + c = bx$

$\bf\implies \:x = \dfrac{ {(y - d)}^{3} }{ {ba}^{3}} + \dfrac{c}{b}$

$\rm :\implies\: \: y \: \in \: R \: as \: x \: is \: polynomial$

$\bf\implies \:Range \: of \: f(x) \: \in \: R$

Additional Information :-

Domain of a function f(x) is defined as set of those values of x for which f(x) is well defined.

Domain of various functions :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf domain \: of \: f(x) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx & \sf R \\ \\ \sf cosx & \sf R \\ \\ \sf logx & \sf x > 0\\ \\ \sf polynomial & \sf R \\ \\ \sf cotx & \sf R - n\pi\\ \\ \sf cosecx & \sf R - n\pi\end{array}} \\ \end{gathered}$

Answer 2

How to find the range of function :

Given :- f(x)

1. Put y = f(x)

2. Solve the equation y = f(x) to get x in terms of y, Let assume that x = g(y)

3. Now, Find the domain of g(y)

4. The values thus obtained for y is the Range of the function f(x).

\large\underline{\sf{Solution-}}

Solution−

Given that,

\rm :\longmapsto\:f(x) = a \sqrt[3]{bx - c} + d:⟼f(x)=a

3

bx−c

+d

To find the range of f(x),

\rm :\longmapsto\:Let \: f(x) = y = a \sqrt[3]{bx - c} + d:⟼Letf(x)=y=a

3

bx−c

+d

\rm :\longmapsto\: y = a \sqrt[3]{bx - c} + d:⟼y=a

3

bx−c

+d

\rm :\longmapsto\: y - d= a \sqrt[3]{bx - c}:⟼y−d=a

3

bx−c

\rm :\longmapsto\: \dfrac{y - d}{a} \: = \: \sqrt[3]{bx - c}:⟼

a

y−d

=

3

bx−c

On cubing both sides, we get

\rm :\longmapsto\: {\bigg(\dfrac{y - d}{a} \bigg) }^{3} = bx - c:⟼(

a

y−d

)

3

=bx−c

\rm :\longmapsto\: {\bigg(\dfrac{y - d}{a} \bigg) }^{3} + c = bx:⟼(

a

y−d

)

3

+c=bx

\bf\implies \:x = \dfrac{ {(y - d)}^{3} }{ {ba}^{3}} + \dfrac{c}{b}⟹x=

ba

3

(y−d)

3

+

b

c

\rm :\implies\: \: y \: \in \: R \: as \: x \: is \: polynomial:⟹y∈Rasxispolynomial

\bf\implies \:Range \: of \: f(x) \: \in \: R⟹Rangeoff(x)∈R

Additional Information :-

Domain of a function f(x) is defined as set of those values of x for which f(x) is well defined.

Domain of various functions :-

\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf domain \: of \: f(x) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx & \sf R \\ \\ \sf cosx & \sf R \\ \\ \sf logx & \sf x > 0\\ \\ \sf polynomial & \sf R \\ \\ \sf cotx & \sf R - n\pi\\ \\ \sf cosecx & \sf R - n\pi\end{array}} \\ \end{gathered}\end{gathered}

f(x)

sinx

cosx

logx

polynomial

cotx

cosecx

domainoff(x)

R

R

x>0

R

R−nπ

R−nπ

I hope it will be help you