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

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

$\displaystyle f'(x)=4x^3+9x^2-4$

$\displaystyle f''(x)=12x^2+18x$

Now we set $\displaystyle f''(x)=0$.

$\displaystyle 12x^2+18x=0$.

Now we factor the left hand side.

$\displaystyle x(12x+18)=0$

From this, we see that there is one point of inflection at $\displaystyle x=0$.

For the point of inflection, lets solve for x for the equation inside the parentheses. 

$\displaystyle \\ 12x+18=0 \\ \\ 12x=-18 \\ \\ x=-\frac{18}{12}=-\frac{3}{2}$

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

$\displaystyle f(x)=x^4-x^2+100$

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

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

Now we set $\displaystyle f''(x)=0$.

$\displaystyle 12x^2-2=0$

$\displaystyle \\ 12x^2=2 \\ \\ x^2=\frac{1}{6} \\ \\ x=\pm\sqrt{\frac{1}{6}}=\pm\frac{1}{\sqrt{6}}$

Thus the points of inflection are  $\displaystyle x=\frac{1}{\sqrt{6}}$ and $\displaystyle x=-\frac{1}{\sqrt{6}}$

Which of the following is a point of inflection on f(x)?

$\displaystyle f(x)=5x^3+x+6$

To find points of inflection, we need to find where the second derivative is 0.

So, find f''(x)

$\displaystyle f(x)=5x^3+x+6$

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

$\displaystyle f''(x)=30x$

So, we have a point of inflection at x=0.

Find f(0) to find the y-coordinate:

$\displaystyle f(0)=5(0)^3+0+6=6$

So our point is at $\displaystyle {}(0,6)$.

Which of the following is a point of inflection of f(x) on the interval $\displaystyle \left[\frac{\pi}{2},\pi\right]$?

$\displaystyle f(x)=5sin(x)+12x$

To find points of inflection, we need to know where the second derivative of the function is equal to zero. So, find the second derivative:

$\displaystyle f(x)=5sin(x)+12x$

$\displaystyle f'(x)=5cos(x)+12$

$\displaystyle f''(x)=-5sin(x)$

So, where on the given interval does $\displaystyle f''(x)=0$?

Well, we know from our unit circle that $\displaystyle sin(\pi)=0$,

So we would have a point of inflection at $\displaystyle x=\pi$, but we still need to find the y-coordinate of our POI. find this by finding $\displaystyle f(\pi)$

$\displaystyle f(\pi)=5sin(\pi)+12(\pi)=12\pi$

So our POI is:

$\displaystyle (\pi,12\pi)$

For a graph to have an inflection point, the second derivative must be equal to zero. We also want the concavity to change at that point. 

$\displaystyle \frac{d^2}{dx^{2}}[x]=0$, for all real numbers, but this is a line and has no concavity associated with it, so not this one.

$\displaystyle \frac{d^2}{dx^{2}}[e^{x}]=e^{x}\neq0$, there are no real values of $\displaystyle x$ for which this equals zero, so no inflection points.

$\displaystyle \frac{d^2}{dx^{2}}[e^{-x}]=e^{-x}\neq0$, same story here.

$\displaystyle \frac{d^2}{dx^{2}}[\ln(x)]=\frac{-1}{x^{2}}\neq0$, so no inflection points here.

This leaves us with 

$\displaystyle f(x)=\frac{10}{1+100e^{-x}}$, whose derivatives are a bit more difficult to take.

$\displaystyle f(x)=10(1+100e^{-x})^{-1}$, so by the chain rule we get$\displaystyle f'(x)=-10(1+100e^{-x})^{-2}(-100e^{-x})=1000e^{-x}(1+100e^{-x})^{-2}$

$\displaystyle f''(x)=-1000e^{-x}(1+100e^{-x})^{-2}-2000e^{-x}(1+100e^{-x})^{-3}(-100e^{-x})=-1000e^{-x}(\frac{1+100e^{-x}-200e^{-x}}{(1+100e^{-x})^{3}})=-1000e^{-x}(\frac{1-100e^{-x}}{(1+100e^{-x})^{3}})$

So, $\displaystyle f''(x)=0$ when $\displaystyle 1-100e^{-x}=0$. So 

$\displaystyle -100e^{-x}=-1\\ e^{-x}=1/100\\ \frac{1}{e^{x}}=\frac{1}{100}\\ e^{x}=100 x=\ln(100)$.  This is our correct answer.

To find the inflection point we must find where the second derivative of a function is 0.

Calculating the second derivative is fairly simple. We just need to know that $\displaystyle (x^n)' = n\cdot x^{n-1}$:

The first derivative is 

$\displaystyle {} 15x^2-18x+5$ 

and if we take the derivative once more we get 

$\displaystyle {} 30x-18$.

Setting this equal to zero, we get 

$\displaystyle {} 30x-18 = 0 \rightarrow x= 18/30 = 3/5$

And now all we have to do is plug this value, 3/5, into our original polynomial, to get the answer.

$\displaystyle {} (3/5, -229/25)$

Points of inflection occur when the second derivative changes signs.

$\displaystyle y=3x^5-10x^4+10x^3+7x+1$

$\displaystyle y'=15x^4-40x^3+30x^2+7$

$\displaystyle y''=60x^3-120x^2+60x=60x(x^2-2x+1)=60x(x-1)^2$

The second derivative equals zero at $\displaystyle x=0$ and $\displaystyle x=1$. However, the factor $\displaystyle (x-1)$ has degree two in the second derivative.  This indicates that $\displaystyle x=1$ is a root of the second derivative with multiplicity two, so the second derivative does not change signs at this value.  It only changes signs at $\displaystyle x=0$.  Since $\displaystyle f(0)=1$, the point of inflection is $\displaystyle (0,1)$.

To find the inflection points of a function we need to take the second derivative and find which values make it zero.

To find the first and second derivative we will need to apply the power rule, $\displaystyle \frac{d}{dx}x^n=nx^{n-1}$.

Given, $\displaystyle y=3x^4+4x^3-6x^2-12x+5$

and applying the power rule we get,

$\displaystyle y'=12x^3+12x^2-12x-12$

$\displaystyle y''=12(3x^2+2x-1)$.

Points of inflection happen when the second derivative changes signs.  The quadratic above changes signs at $\displaystyle x=-1$ and $\displaystyle x=\frac{1}{3}$. 

This can be determined by factoring,

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

or by the quadratic formula,

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

The point of inflection can be found by setting the second derivative equal to 0.

The power rule is given by:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Use the power rule twice to find the second derivative.

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

$\displaystyle \frac{d^2}{dx^2}f(x) = 6x$

Set the second derivative equal to $\displaystyle 0$ and solve for $\displaystyle x$.

$\displaystyle 6x = 0$

$\displaystyle x=0$

Find the point of inflection by plugging $\displaystyle x=0$ back into the original equation.

$\displaystyle 0^3 + 0 + 4 = 4$

Therefore, the point of inflection is 

$\displaystyle (0,4)$

To find where the points of inflection occur, we must find the points at which the second derivative changes sign. 

First, we must find the second derivative:

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

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

The derivatives were found using the following rule:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, we must find the x values at which the second derivative is equal to zero:

$\displaystyle x=0$

Now, we make the intervals on which we will examine the sign of the second derivative, using the above x value:

$\displaystyle (-\infty, 0), (0, \infty)$

Note that at the bounds of the intervals the second derivative is neither positive nor negative.

On the first interval, the second derivative is negative, while on the second, the second derivative is positive. (We check the sign of the second derivative by plugging in any value on the interval into the second derivative function.) A sign change occured at $\displaystyle x=0$, so this is our answer.