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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?
Quote:
Originally Posted by Baatata
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 10/32Quote:
Originally Posted by twostep08
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?)