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### Combination Of Functions

Learn Intro to Composing Functions & fg x, practice Composition of Functions Example.

# Combination of Functions

Let us say, we have two functions 'f' and 'g' ,then we can make new functions from these by suitable combinations. Four such combinations are

(1) $$f + g$$              (3) $$f g$$

(2) $$f - g$$               (4) $$\dfrac {f}{g }$$

The definition is similar to the way we define addition, subtraction, division & multiplication in real numbers.

So

$$(f + g) (x) = f(x) + g(x)$$                    $$fg(x) = f(x)\; g(x)$$

$$(f – g) (x) = f(x) – g(x)$$                         $$\Big(\dfrac {f}{g}\Big)(x) = \Large \frac {f(x)}{g(x)}$$

#### Let $$f(x) = x^2 + 2x$$ and $$g(x) = 4x – 1$$  then $$(f + g)(x)$$  will be

A $$x^2 + 8x$$

B $$7x^2 + 9x - 10$$

C $$x^2 + 6x - 1$$

D $$x^2 + x$$

×

$$(f + g) (x) = f(x) + g(x)$$

$$=(x^2 + 2x) + (4x - 1)$$

$$= x^2 + 2x + 4x - 1$$

$$= x^2 + 6x - 1$$

### Let $$f(x) = x^2 + 2x$$ and $$g(x) = 4x – 1$$  then $$(f + g)(x)$$  will be

A

$$x^2 + 8x$$

.

B

$$7x^2 + 9x - 10$$

C

$$x^2 + 6x - 1$$

D

$$x^2 + x$$

Option C is Correct

#### Let $$f(x) = 3x+2$$ and $$g(x)=x^2-5x+1$$, then the value of $$\left ( \dfrac {f}{g} \right)(2)$$ is

A $$\dfrac {5}{8}$$

B $$\dfrac {-8}{5}$$

C $$\dfrac {7}{3}$$

D $$\dfrac {-3}{7}$$

×

$$\left (\dfrac {f}{g}\right)(x)=\dfrac {f(x)}{g(x)}$$  where  $$g(x)\neq0$$

In this case, $$f(x)=3x+2,\,\;g(x)=x^2-5x+1$$

$$\therefore \;\left (\dfrac {f}{g}\right)(2)=\dfrac {f(2)}{g(2)}=\dfrac {3×2+2}{2^2-5×2+1}=\dfrac {8}{-5}=\dfrac {-8}{5}$$

### Let $$f(x) = 3x+2$$ and $$g(x)=x^2-5x+1$$, then the value of $$\left ( \dfrac {f}{g} \right)(2)$$ is

A

$$\dfrac {5}{8}$$

.

B

$$\dfrac {-8}{5}$$

C

$$\dfrac {7}{3}$$

D

$$\dfrac {-3}{7}$$

Option B is Correct

#### Let $$f(x)=\sqrt {2-x}$$ and $$g(x)=\sqrt {x^2-1}$$, then value of $$(f\,g)\,(1)$$ is

A 0

B 7

C 8

D –2

×

$$(fg)\,(1)=f(1)\,g(1)$$ (by definition)

$$=\sqrt {2-1}×\sqrt {1^2-1}$$  $$=\sqrt 1 ×\sqrt {1-1}$$

$$=\sqrt 1 ×\sqrt {0}$$

$$=1×0=0$$

### Let $$f(x)=\sqrt {2-x}$$ and $$g(x)=\sqrt {x^2-1}$$, then value of $$(f\,g)\,(1)$$ is

A

0

.

B

7

C

8

D

–2

Option A is Correct

# Combination Functions (Product)

• The product function of two function $$f$$ and $$g$$ is defined as $$(fg)\,(x)=f(x)\,g(x)$$
• If this product function is to be found at particular value of $$x$$ say $$u=\alpha$$, then $$(fg)\,(\alpha)=f(\alpha)\,g(\alpha)$$. So find $$f(\alpha)$$ and $$g(\alpha)$$ separately and obtain the product.

#### Let $$f(x)=\sqrt {2+x}$$ and $$g(x)=(2x^2+7)$$, find $$(f\,g)\,(x)$$ .

A $$\sqrt {3x+5}\,(x^2+2)$$

B $$(2x^2+7)\;\sqrt {2+x}$$

C $$\dfrac {3+x^2}{x}$$

D $$\dfrac {x+7}{9}$$

×

$$(fg)\,(x)=f(x)\,g(x)$$ (by definition)

$$f\,(g(x))=\sqrt {2+x}×(2x^2+7)$$

$$f\,(g(x))=(2x^2+7)\;\sqrt {2+x}$$

### Let $$f(x)=\sqrt {2+x}$$ and $$g(x)=(2x^2+7)$$, find $$(f\,g)\,(x)$$ .

A

$$\sqrt {3x+5}\,(x^2+2)$$

.

B

$$(2x^2+7)\;\sqrt {2+x}$$

C

$$\dfrac {3+x^2}{x}$$

D

$$\dfrac {x+7}{9}$$

Option B is Correct

#### Let $$f(x) = 4x^2+x+1$$ and $$g(x)=2x-1$$, then the value of $$\left ( f+g \right)(-5)$$ is

A 17

B 85

C 92

D –19

×

$$\left (f+g\right)(x)=f(x)+g(x)$$

In this case, $$f(x)=4x^2+x+1,\,\;g(x)=2x-1$$

$$\therefore \; (f+g)(-5)=f(-5)+g(-5)=(4×(-5)^2-5+1)+(2×(-5)-1)$$

$$=96+(-11)=85$$

$$\therefore \;(f+g)(-5)=85$$

### Let $$f(x) = 4x^2+x+1$$ and $$g(x)=2x-1$$, then the value of $$\left ( f+g \right)(-5)$$ is

A

17

.

B

85

C

92

D

–19

Option B is Correct

# Combination of Function (Division)

• The division function of two $$f$$ function and $$g$$ is defined as

$$h(x)=\dfrac {f}{g}(x)=\dfrac {f(x)}{g(x)}\rightarrow$$ by definition

• If this division function is to be found at a particular value of  $$x$$ say $$x=\alpha$$ ,then

$$(f \,/g)(\alpha)=f(\alpha)/g(\alpha)$$

so find $$f(\alpha)$$ and $$g(\alpha)$$ separately and obtain the division.

#### Let $$f(x)=2x^2+3x+1$$ and $$g(x)=x+2$$, then the function  $$h(x)=\dfrac {f}{g}(x)$$ will be

A $$\dfrac {5x+3}{x+7}$$

B $$\dfrac {2x^2+3x+1}{x+2}$$

C $$\dfrac {x-2}{x^2+1}$$

D $$\dfrac {\sqrt {x+3}}{x+6}$$

×

$$h(x)=\dfrac {f}{g}(x)=\dfrac {f(x)}{g(x)}\rightarrow$$ by definition

$$=\dfrac {2x^2+3x+1}{x+2}$$

### Let $$f(x)=2x^2+3x+1$$ and $$g(x)=x+2$$, then the function  $$h(x)=\dfrac {f}{g}(x)$$ will be

A

$$\dfrac {5x+3}{x+7}$$

.

B

$$\dfrac {2x^2+3x+1}{x+2}$$

C

$$\dfrac {x-2}{x^2+1}$$

D

$$\dfrac {\sqrt {x+3}}{x+6}$$

Option B is Correct