Function

Here,
$\begin{aligned} \text{Given, } \text{f(x)} & \text{ = } \text{2x - 1} \\ \text{ff(-1)} & \text{ = } \text{?} \\ \end{aligned}$

Now,
$\begin{aligned} \text{ff(x)} & \text{ = } \text{f(2x - 1)} \\ & \text{ = } \text{2(2x - 1) - 1} \\ & \text{ = } \text{4x - 3} \\ \therefore \text{ff(-1)} & \text{ = } \text{-4 - 3} \\ & \text{ = } \text{-7} \end{aligned}$

Here,
$\begin{aligned} \text{Given, } \text{f(x)} & \text{ = } \text{2x + 3} \\ \text{ff(-1)} & \text{ = } \text{?} \\ \end{aligned}$

Now,
$\begin{aligned} \text{ff(x)} & \text{ = } \text{f(2x + 3)} \\ & \text{ = } \text{2(2x + 3) + 3} \\ & \text{ = } \text{4x + 9} \\ \therefore \text{ff(2)} & \text{ = } \text{8 + 9} \\ & \text{ = } \text{17} \end{aligned}$

Solution,
$\begin{aligned} \text{Given, } \text{h(x)} & \text{ = } \text{2x + 3} \\ \text{g(x)} & \text{ = } \text{3x - 2} \\ \end{aligned}$

Now,
$\begin{aligned} \text{hog(x)} & \text{ = } \text{h(3x - 2)} \\ & \text{ = } \text{2(3x - 2) + 3} \\ & \text{ = } \text{6x - 4 + 3} \\ & \text{ = } \text{6x - 1} \\ \therefore \text{hog(5)} & \text{ = } \text{30 - 1} \\ & \text{ = } \text{29} \end{aligned}$

Solution,
Given, f(x) = y = $\frac{\text{2x + 3}}{2}$
For f^-1(x),

$\begin{aligned} y & = \frac{\text{2x + 3}}{2} \\ \text{or, x} & = \frac{\text{2y - 3}}{2} \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} y & = \frac{\text{2x - 3}}{2} \\ \therefore f^{-1}(x) & = \frac{\text{2x - 3}}{2} \\ \end{aligned}$

Solution,
$\begin{aligned} Given, f(x) & = x + 1 \\ g(x) & = 2x + 1 \\ \end{aligned}$

Now,
$\begin{aligned} \text{gof(x)} & = \text{g(x + 1)} \\ & = \text{2(x + 1) + 1} \\ \therefore \text{gof(x)} & = \text{2x + 3} \\ \end{aligned}$

Solution,
$\begin{aligned} Given, f(x) & = 3x - 2 \\ fog(x) & = 6x - 2 \\ g(x) & = \text{a (let)}\\ \end{aligned}$

Now,
$\begin{aligned} \text{fog(x)} & = \text{6x - 2} \\ \text{or, f(a)} & = \text{6x - 2} \\ \text{or, 3a - 2} & = \text{6x - 2} \\ \text{or, 3a} & = \text{6x} \\ \text{or, a} & = \text{2x} \\ \therefore \text{g(x)} & = \text{2x} \\ \end{aligned}$

Solution,
Given, f(x) = y = 4x + 3
To find, f^-1(4) = ?

Now, for f^-1(x)
$\begin{aligned} \text{y} & = \text{4x + 3} \\ \text{or,} \frac{y - 3}{4} & = \text{x} \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or,} \frac{x - 3}{4} & = \text{y} \\ \text{So, }f^{-1}(x) & = \frac{x - 3}{4} \\ \therefore f^{-1}(4) & = \frac{4 - 3}{4} \\ & = \frac{1}{4} \\ \end{aligned}$

Solution
Given, f^-1(x) = y = 2x - 3
To find, f(x) = ?

Now,
$\begin{aligned} \text{y} & = \text{2x - 3} \\ \text{or,} \frac{y + 3}{2} & = \text{x} \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or,} \frac{x + 3}{2} & = \text{y} \\ \therefore f(x) & = \frac{x + 3}{2} \\ \end{aligned}$

Solution,
Given, f^-1(x) = y = $\frac{\text{x + 3}}{2}$ To find, f(x) = ?

Now,
$\begin{aligned} \text{y} & = \frac{\text{x + 3}}{2} \\ \text{or, 2y - 3} & = \text{x} \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or,2x - 3} & = \text{y} \\ \therefore f(x) & = \text{2x - 3} \\ \end{aligned}$

Solution,
$\begin{aligned} Given, f(x) & = 2x + 5 \\ g(x) & = 3x - 1 \\ \end{aligned}$

Now,
$\begin{aligned} \text{gof(x)} & = \text{g(2x + 5)} \\ & = \text{3(2x + 5) - 1} \\ & = \text{6x + 15 - 1} \\ \therefore \text{gof(x)} & = \text{6x + 14} \\ \end{aligned}$

Here,
$\begin{aligned} Given, f(x) & = 2x + 5 \\ g(x) & = \frac{\text{x - 5}}{2} \\ \end{aligned}$
To prove: gh(x) is an identity function.

Now,
$\begin{aligned} \text{fog(x)} & = f\left(\frac{\text{x - 5}}{2}\right) \\ & = \frac{\text{2(x - 5)}}{2} + 5 \\ & = \text{x} \\ \end{aligned}$

Here, fog(x) = x. So, fog(x) is an identity function.

Here,
$\begin{aligned} Given, g(x) & = \frac{\text{x - 2}}{3} \\ h(x) &= 3x + 2 \\ \end{aligned}$
To prove: gh(x) is an identity function.

Now,
$\begin{aligned} \text{gh(x)} & = g(3x + 2) \\ & = \frac{\text{3x + 2 - 2}}{3} \\ & = \frac{\text{3x}}{3} \\ & = \text{x} \\ \end{aligned}$

Here, gh(x) = x. So, gh(x) is an identity function.

Solution,
$\begin{aligned} \text{Given, } \text{f(x)} & = \text{2x - 3} \\ \text{g(x)} & = x^2\text{ + 1} \\ \end{aligned}$

Now,
$\begin{aligned} \text{fog(x)} & = f(x^2\text{ + 1)} \\ & = 2(x^2\text{ + 1) - 3} \\ & = 2x^2\text{ + 2 - 3} \\ & = 2x^2\text{ - 1} \\ \therefore \text{fog(3)} & = 2(3)^2 - 1 \\ & = 18 - 1 \\ & = 17 \\ \end{aligned}$

Solution,
Given, g(x) = y = 4x - 2
To find, g^-1(-1) = ?

Now, for g^-1(x)
$\begin{aligned} \text{y} & = \text{4x - 2} \\ \text{or, y + 2} & = \text{4x} \\ \text{or,} \frac{y + 2}{4} & = \text{x} \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or,} \frac{x + 2}{4} & = \text{y} \\ \text{So, }g^{-1}(x) & = \frac{x + 2}{4} \\ \therefore g^{-1}(-1) & = \frac{-1 + 2}{4} \\ & = \frac{1}{4} \\ \end{aligned}$

Solution,
Given, f = {(1, 2), (2, 3), (3, 4)}, g = {(2, a), (4, c), (3, b)}
Now, gof in arrow diagram, Finally, gof in ordered pair form:
gof = {(1, a), (2, b), (3, c)}

Solution,
$\begin{aligned} \text{Given, } f(x) & = 3x \\ g(x) & = x + 2 \\ fog(x) &= 18 \\ \end{aligned}$
To find: x = ?,

Now,
$\begin{aligned} \text{fog(x)} & = 18 \\ \text{or, f(x + 2)} & = 18 \\ \text{or, 3(x + 2)} & = 18 \\ \text{or, x + 2} & = 6 \\ \therefore \text{x} & = 4 \\ \end{aligned}$

Solution,
$\begin{aligned} \text{Given, } \text{f(x)} & = \text{3x + b} \\ \text{ff(2)} & = \text{12}\\ \end{aligned}$
To find: b = ?,

Now,
f(x) = 3x + b
f(2) = 6 + b ---- (A)
Again,
$\begin{aligned} \text{ff(2)} & = \text{12} \\ \text{or, f(6 + b)} & = \text{12 [from (A)]} \\ \text{or, x + 2} & = \text{6} \\ \therefore \text{x} & = \text{4} \\ \end{aligned}$

Solution,
Given, f(x) = 4x - 5
Range of f(x) = (-1, 7)
To find, Domain of f(x) = ?,

Now,
f(x) = 4x - 5
Put f(x) = -1
So, -1 = 4x - 5
or, 4 = 4x
∴ x = 1
Again, put f(x) = 7
12 = 4x
∴ x = 3
Hence, domain of f(x) = (1, 3).