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A function $f$ is such that $f^{\prime}(x)=6-4 \sin 2 x$ and $f(0)=3$. What is $f(x)$ equal to?
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Verified Answer
The correct answer is:
$6 x+2 \cos 2 x+1$
Given,
$f^{\prime}(x)=6-4 \sin 2 x$ and $f(0)=3$
Consider $f^{\prime}(x)=6-4 \sin 2 x$
Integrate both sides w.r.t $x$
$\int f^{\prime}(x) d x=\int(6-4 \sin 2 x) d x$
$f(x)=6 x-\frac{4(-\cos 2 x)}{2}+c$
Where ' $c$ ' is constant of integration $\Rightarrow f(x)=6 x+2 \cos 2 x+c$
By using $f(0)=3$, we have $3=f(0)=6.0+2 \cos 0+c$
$\Rightarrow 3=2+c \Rightarrow c=1$
Hence, $f(x)=6 x+2 \cos 2 x+1$
$f^{\prime}(x)=6-4 \sin 2 x$ and $f(0)=3$
Consider $f^{\prime}(x)=6-4 \sin 2 x$
Integrate both sides w.r.t $x$
$\int f^{\prime}(x) d x=\int(6-4 \sin 2 x) d x$
$f(x)=6 x-\frac{4(-\cos 2 x)}{2}+c$
Where ' $c$ ' is constant of integration $\Rightarrow f(x)=6 x+2 \cos 2 x+c$
By using $f(0)=3$, we have $3=f(0)=6.0+2 \cos 0+c$
$\Rightarrow 3=2+c \Rightarrow c=1$
Hence, $f(x)=6 x+2 \cos 2 x+1$
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