Question

For the function $y=x^3+21$,find the value of $y$ increases $75$ times as fast as $x$

Answer 1

$y=x^3+21$

$\therefore \frac{dy}{dx}=\frac{d(x^3+21)}{dx}=3x^2$

Now value of $y$ increses $75$ times that of $x$

$\therefore \frac{dy}{dx}=75$

Thus,$3x^2=75$

$x^2=25$

$x=\pm 5$

Thus $y=5^3+21=146$ or $y=(-5{)}^3+21=-104$