We can determine the production level that maximizes profit by taking the derivative of our function.  This can be done using the power rule.

$\displaystyle \small p'(x)=3(-2000x^2)+2(12000x)+126000=-6000x^2+24000x+126000$

We then set this derivative equal to 0 and solve.

$\displaystyle \small 0=-6000x^2+24000x+126000$

We can factor out our greatest common factor and then divide it out.

$\displaystyle \small 0=-6000(x^2-4x-21)$

$\displaystyle \small 0=x^2-4x-21$

Next we can factor.

$\displaystyle \small 0=(x+3)(x-7)$

Therefore, we can solve

$\displaystyle \small x=-3\:or\:7$

We remember that these production levels are in thousands of units.

However, a factory cannot produce negative cellphones, so the maximum profit is obtained when seven thousand units are produced.

To find the critical points, we first take the first derivative using the Power Rule:

$\displaystyle y=x^n \rightarrow y'=nx^{n-1}$

Therefore,

$\displaystyle y=x^2$

$\displaystyle y'=2x$. 

To find the critical point we need to set the derivative equal to zero and solve for x.

Doing so this gives the critical point at $\displaystyle x=0$. To get the y value of the point plug x=0 back into the original equation.

$\displaystyle y=0^2=0$ thus the point is $\displaystyle (0,0)$.

Is it a max or min?

Take the second derivative

$\displaystyle y''=2$

It is a minimum since the second derivative is positive.

To find the critical points, we first take the first derivative using the Power Rule.

$\displaystyle y=x^n \rightarrow y'=nx^{n-1}$.

Therefore,

$\displaystyle y=-x^2$

$\displaystyle y'=-2x$. 

From here we need to set the derivative equal to zero and solve to x.

The critical point occurs when $\displaystyle x=0$. Now to get the y value of the point we will need to plug in x=0 into the original equation.

$\displaystyle y=-0^2=0$. This gives us a critical point at $\displaystyle (0,0)$.

To find if it is a max or min, we take the second derivative

$\displaystyle y''=-2$

Since the second derivative is negative, it is a maximum.

Determine the derivative of this function.

$\displaystyle y=3x^2-11x+5$

$\displaystyle y'=6x-11$

Set the derivative function equal to zero and solve for $\displaystyle x$.

$\displaystyle 0=6x-11$

$\displaystyle x=\frac{11}{6}$

Since the parabola has a positive coefficient for $\displaystyle 3x^2$, the parabola will open upwards, and therefore will have a minimum.

Substitute $\displaystyle x=\frac{11}{6}$back into the original function and find the y-value of the minimum.

$\displaystyle y=3(\frac{11}{6})^2-11(\frac{11}{6})+5$

$\displaystyle = 3(\frac{121}{36})-\frac{121}{6}+5$

$\displaystyle =\frac{121}{12}-\frac{121}{6}+5$

$\displaystyle =\frac{121}{12}-\frac{242}{12}+\frac{60}{12}$

$\displaystyle y=-\frac{61}{12}$

The parabola has a minimum at the point $\displaystyle (\frac{11}{6},-\frac{61}{12})$.

Let's find the first derivative to locate the relative maxima and minima.

$\displaystyle y=\frac{3}{4}x^4+3x^3-15x^2+8$

$\displaystyle y'=4*\frac{3}{4}x^3+9x^2-30x+0$

$\displaystyle y'=3x^3+9x^2-30x$

Now we set it equal to zero to find the x values of these critical points.

$\displaystyle 0=3x^3+9x^2-30x$

$\displaystyle 0=3x(x^2+3x-10)$

$\displaystyle 0=3x(x-2)(x+5)$

So the equation is 0 where x is -2, 0, or 5. Now let's find the second derivative so that we know which of these locations are maxima and which are minima.

$\displaystyle y'=3x^3+9x^2-30x$

$\displaystyle y''=9x^2+18x-30$

$\displaystyle y''(-5)=9(-5)^2+18(-5)-30=105>0$

So the function has a relative maximum at x=-5.

$\displaystyle y''(0)=9(0)^2+18(0)-30=-30< 0$

So the function has a relative minimum at x=0.

$\displaystyle y''(2)=9(2)^2+18(2)-30=42>0$

So the function has a relative maximum at x=2.

Thus there is only one relative minimum in this function, and it occurs at x=0. We need to plug this into the original function to find the y-coordinate of the point.

$\displaystyle y=\frac{3}{4}x^4+3x^3-15x^2+8$

$\displaystyle y=\frac{3}{4}(0)^4+3(0)^3-15(0)^2+8$

$\displaystyle y=0+0-0+8$

$\displaystyle y=8$

So our point is (0,8).

We need to find out where the first derivative is equal to zero to find the locations of the possible maxima and minima.

$\displaystyle f(x)=\frac{1}{2}x^4-x^2+3$

$\displaystyle f'(x)=\frac{4}{2}x^3-2x+0$

$\displaystyle f'(x)=2x^3-2x$

$\displaystyle f'(x)=2x(x^2-1)$

$\displaystyle f'(x)=2x(x+1)(x-1)$

$\displaystyle 0=2x(x+1)(x-1)$

$\displaystyle x=0,1,-1$ or $\displaystyle x=0,\pm1$.

We see that the function is a quadratic, so its graph will be a parabola. Thus, we know it will have only one relative max or min. Since the leading coefficient of $\displaystyle y=6x^2-2x+9$ is a positive 6, we also know that the parabola opens upward. Hence, the relative extremum is a minimum.