The points of inflection of a function occur where the second derivative of the funtion is equal to zero.

Find this second derivative by taking the derivative of the function twice:

 $\displaystyle f(x)=x^3+21x^2+10$

$\displaystyle f'(x)=3x^2+42x$

$\displaystyle f'(x)=6x+42$

Set the second derivative to zero and find the values that satisfy the equation:

 $\displaystyle 6x+42=0$

$\displaystyle x=-7$

It can be shown that this is a point of inflection by the change in sign of the second derivative with points before and after this value

$\displaystyle 6(-6)+42=6$

$\displaystyle 6(-8)+42=-6$

Now, plug this value back in to the original function to find the value of the function that matches:

$\displaystyle f(x)=(-7)^3+21(-7)^2+10=696$

The point of inflection is $\displaystyle (-7,696)$

It can be confirmed that $\displaystyle x=-7$ is a point of inflection due to the sign change around this point. Picking a greater and lower value $\displaystyle x=-8,0$, observe the difference in sign of the second derivative:

$\displaystyle f''(-8)=6(-8)+42=-6$

$\displaystyle f''(0)=6(0)+42=42$

The points of inflection of a function occur where the second derivative of the funtion is equal to zero.

Find this second derivative by taking the derivative of the function twice:

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

To take these derivatives, make use of the following rules:

$\displaystyle d[sin(u)]=cos(u)du$

$\displaystyle d[cos(u)]=-sin(u)du$

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

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

Set the second derivative to zero and find the values that satisfy the equation:

 $\displaystyle 2x-2\pi cos(x)=0$

$\displaystyle x=-\pi,-2.48,1.18$

These can be shown to be points inflection by observing how the signs on the plot of $\displaystyle f''(x)=2x-2\pi cos(x)$; note how the function crosses the x-axis at these values

Now, plug these values back in to the original function to find the values of the function that match to them:

$\displaystyle f(-\pi)=\frac{1}{3}(-\pi)^3+2\pi cos(-\pi)=-16.62$

$\displaystyle f(-2.48)=\frac{1}{3}(-2.48)^3+2\pi cos(-2.48)=-10.04$

$\displaystyle f(1.18)=\frac{1}{3}(1.18)^3+2\pi cos(1.18)=2.94$

The points of inflection are $\displaystyle (-\pi,-16.62)$, $\displaystyle (-2.48,-10.04)$, $\displaystyle (1.18,2.94)$

First, find the derivative using the power rule:

Remember that the power rule is:

$\displaystyle (x^n)'=nx^{n-1}$

Apply this to our problem to get

$\displaystyle 14x+1$

Now, substitute $\displaystyle 3$ for $\displaystyle x$.

$\displaystyle 14(3)+1=43$

First, use the power rule to find the derivative.

Remember that the power rule is:

$\displaystyle (x^n)'=nx^{n-1}$

Apply this to our problem to get

$\displaystyle 2+x$

Now, substitute $\displaystyle 6$ for $\displaystyle x$.

$\displaystyle 2+6=8$

Find the derivative using the power rule.

Remember that the power rule is:

$\displaystyle (x^n)'=nx^{n-1}$

Apply this to our problem to get

$\displaystyle 4x^3+6x^2$

Now, substitute $\displaystyle 0$ for $\displaystyle x$. 

$\displaystyle 4(0)^3+6(0)^2$

$\displaystyle 0$

First, find the derivative using the power rule. 

Remember that the power rule is:

$\displaystyle (x^n)'=nx^{n-1}$

Apply this to our problem to get

$\displaystyle 15x^2+4x-1$

Now, substitute $\displaystyle 1$ for $\displaystyle x$.

$\displaystyle 15(1)^2+4(1)-1=15+4-1=18$

Use the power rule to find this derivative.

Remember that the power rule is:

$\displaystyle (x^n)'=nx^{n-1}$

Apply this to our problem to get

$\displaystyle \frac{d}{dx}7x^3=21x^2$

$\displaystyle \frac{d}{dx}x^2=2x$

$\displaystyle \frac{d}{dx}3x=3$

Thus, the derivative is $\displaystyle 21x^2+2x+3$.

The points of inflection of a function occur where the second derivative of the funtion is equal to zero.

Find this second derivative by taking the derivative of the function twice:

 $\displaystyle f(x)=x^4+16x^3+90x^2-50$

$\displaystyle f'(x)=4x^3+48x^2+180x$

$\displaystyle f''(x)=12x^2+96x+180$

Set the second derivative to zero and find the values that satisfy the equation:

$\displaystyle 12x^2+96x+180=0$

$\displaystyle 12(x^2+8x+15)$

$\displaystyle 12(x+3)(x+5)$

$\displaystyle x=-3,-5$ 

These points can be shown to be points of inflection by how the sign changes at points just adjacent to them:

For $\displaystyle x=-3$

$\displaystyle 12(-2)^2+96(-2)+180=36$

$\displaystyle 12(-4)^2+96(-4)+180=-12$

For $\displaystyle x=-5$

$\displaystyle 12(-4)^2+96(-4)+180=-12$

$\displaystyle 12(-6)^2+96(-6)+180=36$

Now, plug these values back in to the original function to find the values of the function that match to them:

$\displaystyle f(-3)=(-3)^4+16(-3)^3+90(-3)^2-50=409$

$\displaystyle f(-5)=(-5)^4+16(-5)^3+90(-5)^2-50=825$

The two points of inflection are

$\displaystyle (-3,409)$, $\displaystyle (-5,825)$

To verify an inflection point, plug a x value higher and lower into the second derivative to find if there is a sign change. If a sign change occurs around the inflection point than it is in fact a true inflection point. To check $\displaystyle x=-5$, plug in a value higher and a value lower than it.

$\displaystyle f''(-6)=12(-6)^2+96(-6)+180=36$

$\displaystyle f''(-4)=12(-4)^2+96(-4)+180=-12$

Use the product rule to find the derivative.

Remember that the product rule is:

$\displaystyle (f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$

Apply this to our problem to get

$\displaystyle \frac{d}{dx}x^2\sin(x)=x^2\cos(x)+2x\sin (x)$

The points of inflection of a function occur where the second derivative of the funtion is equal to zero.

Find this second derivative by taking the derivative of the function twice:

 $\displaystyle f(x)=4x^3-48x^2+420$

$\displaystyle f'(x)=12x^2-96x$

$\displaystyle f''(x)=24x-96$

Set the second derivative to zero and find the values that satisfy the equation:

 $\displaystyle 24x-96=0$

$\displaystyle x=4$

This can be shown to be a point of inflection by how the sign changes for $\displaystyle f''(x)=24x-96$ on either side of it:

$\displaystyle f''(3)=24(3)-96=-24$

$\displaystyle f''(5)=24(5)-96=24$

Now, plug this value back in to the original function to find the value of the function that matches:

 $\displaystyle f(4)=4(4)^3-48(4)^2+420=-92$

The point of inflection is $\displaystyle (4,-92)$.

To verify this is a true inflection point, plug in x values that are higher and lower than four. If a sign change occurs around four than it is in fact an inflection point.

$\displaystyle f''(0)=24(0)-96=-96$

$\displaystyle f''(5)=24(5)-96=24$