Over which interval(s) is g(x) = x2 + x - 12 positive?

Mathematics

1 answer:

VikaD [51]3 years ago

4 0

Answer:

B. x < -4 and x > 3

Step-by-step explanation:

Factor and set = to 0

$x^{2} + x - 12\\( x + 4)(x - 3)$ = 0

x = - 4 or x = 3 I call these critical values

The two numbers would divide a number line into 3 intervals. Pick a value in one of the intervals and put it in the original expression. If it makes the function positive, then all the values in that interval make the function positive. If the value you picked makes the function negative, then the values in the other intervals will make the function negative. Let's pick the value of 0 and substitute it into the function

We get $0^{2}$ + 0 - 12 = -12 which is not positive. Therefore, all the values between -4 and 3 will make the function negative. So, the values less than -4 or greater than 3 will make the function positive. Therefore, B is the correct answer.

Another way to do this problem is to graph the function and see where the graph is above the x-axis. But, sometimes it is not easy to graph the function.

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Use cylindrical coordinates. find the volume of the solid that is enclosed by the cone z = x2 + y2 and the sphere x2 + y2 + z2 =

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Let $R$ be the solid. Then the volume is

$\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

which follows from the facts that

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$
(by computing the Jacobian)

and

$z=x^2+y^2=r^2\implies z+z^2=72\implies z=-9\text{ or }z=8$
(we take the positive solution, since it's clear that $R$ lies above the $x$ - $y$ plane)
$r^2+z^2=72\implies z=\pm\sqrt{72-r^2}$
(again, taking the positive root for the same reason)
$z=r^2\implies 8=r^2\implies r=\sqrt8$

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}r(\sqrt{72-r^2}-r^2)\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr$
$=\dfrac{32(27\sqrt2-35)\pi}3$