Inflection Points question

• Nov 12th 2009, 10:17 AM
Baatata
Inflection Points question
f(x)=-8x^4-5x^3 +3 for concavity and inflection points.

f(x)=-8x^4-5x^3 +3
f'(x)=-32x^3-15x^2
f''(x)=-96x^2-30x

can someone explain to me what exactly Inflection Points are?
is it like critical points for the first derivative but those are for the second one?
i mean, making f''(x) = 0 and get 2 numbers and those are the Inflection Points?
• Nov 12th 2009, 11:11 AM
twostep08
Quote:

Originally Posted by Baatata
f(x)=-8x^4-5x^3 +3 for concavity and inflection points.

f(x)=-8x^4-5x^3 +3
f'(x)=-32x^3-15x^2
f''(x)=-96x^2-30x

can someone explain to me what exactly Inflection Points are?
is it like critical points for the first derivative but those are for the second one?
i mean, making f''(x) = 0 and get 2 numbers and those are the Inflection Points?

inflection points are when the original function changes concavity.
so, yes, set the second derivative equal to 0, and see where the changes occur
• Nov 12th 2009, 11:52 AM
Baatata
Quote:

Originally Posted by twostep08
inflection points are when the original function changes concavity.
so, yes, set the second derivative equal to 0, and see where the changes occur

okay so i get x=0 and 10/32
is this right?

now i have to plug it in f(x) and those are my 2 values? (if yes then i get 3 and 2.923 is this right?)