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The function $f(x)=3 x^{4}+16 x^{3}-30 x^{2}+10$ is increasing for
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Verified Answer
The correct answer is:
$x \in(-5,0) \cup(1, \infty)$
$f(x)=3 x^{4}+16 x^{3}-30 x^{2}+10$
$\therefore f^{\prime}(x)=12 x^{3}+48 x^{2}-60 x$
When $f^{\prime}(x)>0$, we write
$\quad x\left(12 x^{2}+48 x-60\right)>0$
$12 x\left(x^{2}+4 x-5\right)>0$
$\therefore f^{\prime}(x)>0$, when $x \in(-5,0) \cup(1, \infty)$
$\therefore f^{\prime}(x)=12 x^{3}+48 x^{2}-60 x$
When $f^{\prime}(x)>0$, we write
$\quad x\left(12 x^{2}+48 x-60\right)>0$
$12 x\left(x^{2}+4 x-5\right)>0$
$\therefore f^{\prime}(x)>0$, when $x \in(-5,0) \cup(1, \infty)$
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