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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

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

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

Illustration Questions

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

Illustration Questions

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

Illustration Questions

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

Illustration Questions

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.

Illustration Questions

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

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.

 

Illustration Questions

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

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