0
$\begingroup$

I am unsure where the inequality x>0 comes into play in this question. Is it just there to confuse me? I differentiated the equation to get $12x^2+x^{-0.5}$. Is this the final answer or do I have to do something with the given inequality?

$\endgroup$
1
$\begingroup$

yes it is $$y'(x)=12x^2+2\cdot \frac{1}{2}x^{-1/2}$$

$\endgroup$
1
$\begingroup$

The inequality is only there because $x^{\frac{1}{2}}$ only exists for $x \geq 0$, and once you differentiate that you get, as you correctly found, $\frac{1}{2}x^{\frac{-1}{2}}$, which only exists for $x > 0$. So you've done it correctly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.