Points of inflection

[blehmag]

i need help with starting off these two problems

y=tan^-1x

y=xabs9-x^2

help would be really appreciated

 

[daon]

Take the second derivative and set it equal to zero and solve for x (or y, t, whatever -- and there may be several values). This point may either be a stationary inflection point (f'(x) is zero; i.e. x is also a critical point) or a standard inflection point.

 

[blehmag]

would the derivative of the first problem be

-sec^2x, or am i completely lost?

 

[daon]

the derivative of tanx is sec^2x. (not what you want)

you have f(x)=arctan(x).

 

[blehmag]

oo, okay, i get it now,

this other problem y=ex^x

i had gotten a o and a 1 for points of inflection but, i checked the answer and it said two?
im confused, can you use the product rule to solve for this?

 

[daon]

For the problem you posted above, y=xe^x, take the second derivative and you'll get a factor of (x-2). Make sure to factor e^x out of everything and simplify as much as possible to get this result.

edit: yes, the product rule

 

[blehmag]

thank you