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Let $g(x)=x^{3}-4 x+6$. If $f^{\prime}(x)=g^{\prime}(x)$ and $f(1)=2$, then what is $f(x)$ equal to?
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Verified Answer
The correct answer is:
$x^{3}-4 x+5$
Now, (gof)' $(x)=1$ Given, $g(x)=x^{3}-4 x+6$
$\operatorname{But} f^{\prime}(x)=g^{\prime}(x)$
$\Rightarrow \int f^{\prime}(x) d x=\int g^{\prime}(x) d x$
$\Rightarrow f(x)=g(x)+c$
$\therefore f(x)=x^{3}-4 x+6+c$ where ' $\mathrm{c}$ ' is a constant. Now, $f(1)=2$ $\Rightarrow 2=f(1)=(1)^{3}-4(1)+6+c$
$\Rightarrow 2=1-4+6+c$
$\Rightarrow c=-1$
$\therefore f(x)=x^{3}-4 x+6-1=x^{3}-4 x+5$
$\operatorname{But} f^{\prime}(x)=g^{\prime}(x)$
$\Rightarrow \int f^{\prime}(x) d x=\int g^{\prime}(x) d x$
$\Rightarrow f(x)=g(x)+c$
$\therefore f(x)=x^{3}-4 x+6+c$ where ' $\mathrm{c}$ ' is a constant. Now, $f(1)=2$ $\Rightarrow 2=f(1)=(1)^{3}-4(1)+6+c$
$\Rightarrow 2=1-4+6+c$
$\Rightarrow c=-1$
$\therefore f(x)=x^{3}-4 x+6-1=x^{3}-4 x+5$
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