Math, asked by shahjungsudhir, 5 months ago

14. a) If f(x + 3)=x + 6, find f '(x).
b) If f(2x + 5)=4x + 13, find f'(x).
c) If f(3x + 4)=5x + 8, find f(x).​

Answers

Answered by arvindrathore2405
0

Answer:

please write down full question

Answered by PharohX
1

Step-by-step explanation:

 \green{ \boxed{  \large\sf \: SOLUTION}}

 \large{ \pink{ \star}}  \sf \large (a) \green{ \rightarrow}

 \sf    f(x + 3) \: =x + 6  \\

 \sf put \:  \: x = x - 3 \: in \: both \: side

 \sf \: we \: get \:  \rightarrow

 \sf \: f(x - 3 + 3) = x - 3 + 6 \:  \\

 \sf \:   \implies \: f(x) =  x - 3  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \\  \sf \:   \implies \:  f '(x)= \frac{d}{dx} (  x - 3 ) \\ \sf \:   \implies \:  f '(x)=  1 - 0 = 1  \: \\ \sf \:   \implies  \green{\boxed{  \sf \: f '(x)=  1 }}\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:

  \large{  \pink\star}\sf \large (b) \green{ \rightarrow}

 \sf    f(2x+ 5) \: =4x + 13  \\

  \displaystyle\sf put \:  \: x =  \frac{x - 5}{2}  \:  \: in \: both \: side

 \sf \: we \: get \:  \rightarrow

 \sf    f(2 \bigg( \frac{x - 5}{2} \bigg) + 5) \: =4\bigg( \frac{x - 5}{2} \bigg)+ 13  \\

  \sf \: f(x - 5 + 5) = 2x - 10 + 13

 \sf \: f(x) = 2x  + 3 \\

 \sf \implies \: f(x) = 2x  + 3  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \\\sf \implies f'(x) =  \frac{d}{dx} (2x +3 )

\sf \implies \: f'(x) = 2 + 0 = 2 \\\sf \implies   \green{ \boxed{ \sf \: f'(x) = 2 }}\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

 \large{ \pink{ \star}}  \sf \large (c) \green{ \rightarrow}

 \sf \: f(3x + 4)=5x + 8

 \displaystyle \sf put \:  \: x =  \frac{x - 4}{3}  \: in \: both \: side

 \sf \: we \: get \:  \rightarrow

 \sf    f(3 \bigg( \frac{x - 4}{3} \bigg) + 4) \: =5\bigg( \frac{x - 4}{3} \bigg)+ 8  \\

  \displaystyle \sf \: f(x - 4+ 4) =  \frac{5}{3}(x)  -  \frac{20}{3}  + 13

  \displaystyle \sf \: f(x ) =  \frac{5}{3}(x)  -  \frac{20}{3}  + 13  \:  \\ \\   \implies \sf \: f '(x) =  \frac{d}{dx}  \bigg( \frac{5}{3}(x)  -  \frac{20}{3}  + 13 \bigg) \: \\ \\   \implies \sf \: f '(x) =  \frac{5}{3}  - 0 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\ \implies  \green{ \boxed{\sf \: f '(x) = \:  \frac{5}{3} }} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

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