To find the critical points, set $\displaystyle f{}'(x)=0$ and solve for $\displaystyle x$.

$\displaystyle f(x)= x^3+3x^2+x$

Differentiate:

$\displaystyle f{}'(x)= 3x^2+6x+1$

Set equal to zero:

$\displaystyle 3x^2+6x+1=0$

Solve for $\displaystyle x$ using the quadratic formula: $\displaystyle x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle a=3,\ b=6,\ c=1$

$\displaystyle x= \frac{-6\pm \sqrt{(6)^2-4(3)(1)}}{2(3)}$

$\displaystyle x= \frac{-6\pm \sqrt{36-12}}{6}}$

$\displaystyle x= \frac{-6+\sqrt{24}}{6}}\ and\ x= \frac{-6-\sqrt{24}}{6}}$

$\displaystyle c=-0.18\ and\ c=-1.82$

In order to find the critical points, we must find $\displaystyle \ f'(x)$ and solve for $\displaystyle \ x$

$\displaystyle f(x)=x^3+2x^2+x+4$

$\displaystyle \\ \\ f'(x)=3x^2+4x+1$

Set  $\displaystyle f'(x)=0$

$\displaystyle \\ \\ 3x^2+4x+1=0$

Use the quadratic equation to solve for $\displaystyle \ x$.

Remember that the quadratic equation is as follows.

$\displaystyle \\ \\ x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$, where a,b and c refer to the coefficents in the

equation  $\displaystyle ax^2+bx+c=0$.

In this case, $\displaystyle a=3$, $\displaystyle b=4$, and $\displaystyle c=1$.

After plugging in those values, we get

$\displaystyle \\ \\ x=\frac{-4\pm\sqrt{4^2-4\cdot3\cdot1}}{2\cdot3} =\frac{-4 \pm \sqrt{4}}{6}=\frac{-4 \pm 2}{6}$.

So the critical points $\displaystyle x$ values are:

$\displaystyle \\ \\ c_1=\frac{-4+2}{6}=\frac{-1}{3}$

$\displaystyle \\ \\ c_2=\frac{-4-2}{6}=-1$

In order to find the critical points, we must find $\displaystyle \ f'(x)$ and solve for $\displaystyle \ x$.

$\displaystyle \\ \\ f(x)=5x^3+10x^2+x+100$

$\displaystyle \\ \\ f'(x)=15x^2+20x+1$

Set  $\displaystyle \ f'(x)=0$ 

$\displaystyle \\ \\ 15x^2+20x+1=0$

Use the quadratic equation to solve for $\displaystyle \ x$.

Remember that the quadratic equation is as follows.

 $\displaystyle \\ \\ x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$, where a,b and c refer to the coefficents in the equation  $\displaystyle ax^2+bx+c=0$.

In this case, $\displaystyle a=15$, $\displaystyle b=20$, and $\displaystyle c=1$.

After plugging in those values, we get. 

$\displaystyle {}\\ x=\frac{-20\pm\sqrt{20^2-4\cdot15\cdot1}}{2\cdot15} =\frac{-20 \pm \sqrt{340}}{30} \\ \\ x=\frac{-20 \pm 2\cdot\sqrt{85}}{30}$ 

So the critical points $\displaystyle x$ values are,

$\displaystyle {}\\ \\ c_1=\frac{-20+2\cdot \sqrt{85}}{30}=\frac{-10+\sqrt{85}}{15} \\ \\ c_2=\frac{-20-2\cdot \sqrt{85}}{30}=\frac{-10-\sqrt{85}}{15}$

First we need to find $\displaystyle f'(x)$.

$\displaystyle f(x)=x^3+3x^2+x+5$

$\displaystyle f'(x)=3x^2+6x+1$

Now we set $\displaystyle f'(x)=0.$

$\displaystyle 3x^2+6x+4=0.$

Now we can use the quadratic equation in order to find the critical points.

Remember that the quadratic equation is 

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$,

where a,b,c refer to the coefficients in the equation

$\displaystyle ax^2+bx+c=0$

In this case, a=3, b=6, and c=1. 

$\displaystyle x=\frac{-6\pm\sqrt{(-6)^2-4(3)(1)}}{(2)(3)}$

$\displaystyle x=\frac{-6\pm\sqrt{24}}{6}=\frac{-6\pm2\sqrt{6}}{6}$

Thus are critical points are

$\displaystyle c_1=\frac{-6+2\sqrt{6}}{6}=-1+\frac{\sqrt{6}}{3}$

$\displaystyle c_2=\frac{-6-2\sqrt{6}}{6}=-1-\frac{\sqrt{6}}{3}$

In order to find the critical points, we need to find $\displaystyle f'(x)$ using the power rule $\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$.

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

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

Now we set $\displaystyle f'(x)=0$, and solve for $\displaystyle x$.

$\displaystyle 2x-2=0$

$\displaystyle 2x=2$

$\displaystyle x=1$

Thus $\displaystyle x=1$ is a critical point.

To find the critical point(s) of a function $\displaystyle f(x)$, take its derivative $\displaystyle f'(x)$, set it equal to $\displaystyle 0$, and solve for $\displaystyle x$.

Given $\displaystyle f(x)=2x^{2}-5x+4$, use the power rule

$\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ to find the derivative. Thus the derivative is, $\displaystyle f'(x)=4x-5=0$.

Since $\displaystyle 4x-5=0$:

$\displaystyle 4x=5$

$\displaystyle x=\frac{5}{4}$

The critical point $\displaystyle c$ is $\displaystyle c=\frac{5}{4}$. 

In order to find the critical points, we must find $\displaystyle f'(x)$ using the power rule $\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$.

$\displaystyle f(x)=3x^3-20x^2+4x+99$

$\displaystyle f'(x)=9x^2-40x+4$.

Now we set $\displaystyle f'(x)=0$.

$\displaystyle 9x^2-40x+4=0$

Now we use the quadratic equation in order to solve for $\displaystyle x$.

Remember that the quadratic equation is as follows.

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$,

where a,b,c correspond to the coefficients in the equation 

$\displaystyle ax^2+bx+c=0$.

In this case, a=9, b=-40, c=4.

$\displaystyle x=\frac{40\pm\sqrt{(-40)^2-4(9)(4)}}{2(9)}$

$\displaystyle x=\frac{40\pm\sqrt{1600-144}}{18}$

$\displaystyle x=\frac{40\pm\sqrt{1456}}{18}=\frac{40\pm4\sqrt{91}}{18}$

Then are critical points are:

$\displaystyle c_1=\frac{40+4\sqrt{91}}{18}=\frac{20+2\sqrt{91}}{9}$

$\displaystyle c_2=\frac{40-4\sqrt{91}}{18}=\frac{20-2\sqrt{91}}{9}$

In order to find the critical points, we first need to find $\displaystyle f'(x)$ using the power rule $\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$..

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

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

Now we set $\displaystyle f'(x)=0$.

$\displaystyle 3x^2-2=0$

$\displaystyle 3x^2=2$

$\displaystyle x^2=\frac{2}{3}$

$\displaystyle x=\pm\sqrt{\frac{2}{3}}$

Thus the critical points are at

$\displaystyle x=\sqrt{\frac{2}{3}}$, and

$\displaystyle x=-\sqrt{\frac{2}{3}}$.

To find critical points the derivative of the function must be found. 

$\displaystyle y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

$\displaystyle y'=x^5+3x^4-13x^3-15x^3$

Critical points occur where the derivative equals zero. 

$\displaystyle 0=x^5+3x^4-13x^3-15x^3$

$\displaystyle 0=x^2(x+1)(x+5)(x-3)$

$\displaystyle x=\left \{ {{-5, -1, 0, 3}} \right \}$

To find the point where the graph of $\displaystyle f(x)$ is not changing, we must set the first derivative equal to zero and solve for $\displaystyle x$.

To evaluate this derivate, we need the following formulae:

$\displaystyle \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1, \frac{d}{dx}cx=c$

$\displaystyle \frac{d}{dx}(x^{n}))=nx^{n-1}, \frac{d}{dx}(cx^{n})=ncx^{n-1}$

$\displaystyle f'(x)=2x^{2-1}-12x^{1-1}+0=2x-12$

Now, setting the derivate equal to $\displaystyle 0$ to find where the graph is not changing:

$\displaystyle 2x-12=0 ==> x=6$

Now, to find the corresponding $\displaystyle y$ value, we plug this $\displaystyle x$ value back into $\displaystyle f(x)$:

$\displaystyle f(6)=6^{2}-12(6)+32=36-72+32=-4$

Therefore, the point where $\displaystyle f(x)$ is not changing is $\displaystyle (6,-4)$