Math, asked by rs200167, 6 hours ago

find the average rate of change of the function f(X)=3-sinx over the given interval (0,90)​

Answers

Answered by AestheticSky
33

 \large\maltese \underbrace{ \frak{Concept :-} }

Average Rate of Change over any interval (a,b) is calculated by the following formula :-

 \\   \leadsto\underline{ \boxed{ \sf average =  \frac{f(b) - f(a)}{b - a} }} \bigstar \\

Let's find f(b) and f(a) individually and then substitute them into the formula!

 \large\maltese \underbrace{ \frak{Solution :-} }

 \\    \quad  \dag\underline{ \sf calculating \: f(a), a = 0 :  - }

 \\    \quad  \rightarrow\sf f(x) = 3 - sinx \\

 \\   \quad\rightarrow \sf f(0) = 3 - sin(0) \\

 \\   \quad  \rightarrow\sf f(0) = 3 - 1 \\

 \\  \quad \rightarrow \boxed{ \sf f(0) = 2} \bigstar \\

\\    \quad  \dag\underline{ \sf calculating \: f(b), b = 90 :  - }

Note:- 90 can be written as π/2

  \\  \quad \rightarrow \sf f  \bigg(\dfrac{ \pi}{2}  \bigg) = 3 - sin \bigg(  \frac{\pi}{2} \bigg) \\

 \\  \quad \rightarrow \sf f \bigg( \frac{\pi}{2}  \bigg) = 3 - 1 \\

 \\  \quad \rightarrow \sf f  \bigg(  \frac{\pi}{2} \bigg) = 3 - 1 \\

 \\  \quad \rightarrow \boxed{ \sf f \bigg(  \frac{\pi}{2} \bigg) = 2} \bigstar \\

\\    \quad  \dag\underline{ \sf substituting \: these \: values \: in \: the \: formula :  - }

  \\ \quad \rightarrow \sf average =  \frac{f \bigg(  \dfrac{\pi}{2} \bigg) - f \bigg( 0\bigg)}{b - a}  \\

 \\  \quad \rightarrow \sf average =  \frac{2 - 3}{ \dfrac{\pi}{2} - 0 }  \\

 \\  \quad  \rightarrow \sf average =    \frac{ - 1}{ \dfrac{\pi}{2} }  \\

 \\  \quad \therefore\boxed{ \boxed{\sf average =  \frac{ - 2}{\pi}}}\bigstar

Answered by mathdude500
11

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

Given function is

\rm :\longmapsto\:f(x) =3- sinx \:  \: on \:  \: \bigg[0, \: \dfrac{\pi}{2} \bigg]

So,

\rm :\longmapsto\:f(0) = 3-sin0 = 3

and

\rm :\longmapsto\:f\bigg[\dfrac{\pi}{2} \bigg] =3- sin\bigg[\dfrac{\pi}{2} \bigg] = 3-1=2

We know that,

Average rate of change in function f(x) on the interval [ a, b ] is

\red{ \boxed{ \sf{ \:Average \: rate \: of \: change \:  =  \:  \frac{f(b) - f(a)}{b - a} }}}

So,

\rm :\longmapsto\: \:Average \: rate \: of \: change \:  =  \:  \dfrac{f\bigg[\dfrac{\pi}{2} \bigg] - f(0)}{\dfrac{\pi}{2}  - 0}

\rm :\longmapsto\: \:Average \: rate \: of \: change \:  =  \:  \dfrac{2 - 3}{\dfrac{\pi}{2} }

\rm :\longmapsto\: \:Average \: rate \: of \: change \:  =  \:  \dfrac{-1}{\dfrac{\pi}{2}}

\bf\implies \:\: \:Average \: rate \: of \: change \:  =  \:  \dfrac{-2}{\pi}

Additional Information :-

\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\  \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf  -  \: sinx \\ \\ \sf tanx & \sf  {sec}^{2}x \\ \\ \sf cox & \sf  -  {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf  -  \: cosecx \: cotx\\ \\ \sf  \sqrt{x}  & \sf  \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf  {e}^{x}  & \sf  {e}^{x}  \end{array}} \\ \end{gathered}

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