Question

$\mathtt\red{ if \: f'(x) = 1 - \frac{4}{ {x}^{2} } \: and \: f(1) = 6 \: find \: f(x) \: and \: f(2).}$
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Answer 1

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

Given function is

$\rm :\longmapsto\:f'(x) = 1 - \dfrac{4}{ {x}^{2} }$

On integrating both sides w. r. t. x, we get

$\rm :\longmapsto\:\displaystyle\int\rm f'(x) dx=\displaystyle\int\rm \bigg[ 1 - \dfrac{4}{ {x}^{2} }\bigg] \: dx$

$\rm :\longmapsto\:f(x) = \displaystyle\int\rm 1 \: dx \: - 4\displaystyle\int\rm {x}^{ - 2} \: dx$

We know that

$\boxed{\tt{ \displaystyle\int\rm {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} \: + \: c \: }}$

So, using this identity, we get

$\rm :\longmapsto\:f(x) = x - 4 \bigg[\dfrac{ {x}^{ - 2 + 1} }{ - 2 + 1} \bigg] + c$

$\rm :\longmapsto\:f(x) = x - 4 \bigg[\dfrac{ {x}^{ - 1} }{ - 1} \bigg] + c$

$\bf :\longmapsto\:f(x) = x + \dfrac{4}{x} + c$

Now, Given that

$\red{\rm :\longmapsto\:f(1) = 6}$

$\rm :\longmapsto\:1 + \dfrac{4}{1} + c = 6$

$\rm :\longmapsto\:1 +4 + c = 6$

$\rm :\longmapsto\:5 + c = 6$

$\bf\implies \:c = 1$

So,

$\bf\implies \:f(x) = x + \dfrac{4}{x} + 1$

Now,

$\bf\implies \:f(2) = 2 + \dfrac{4}{2} + 1 = 5$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

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Step-by-step explanation:

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