How do you test #f(x)=8 x^4−9 x^3 +9# for concavity and inflection points?

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Answer 1 · 2018-04-11T19:15:34

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To test for the concavity and inflection points you need to equate the second order derivative with zero.

Keeping in mind:

We proceed:

#f(x)=8x^4-9x^3+9#

#=>f'(x)=32x^3-27x^2#

#=>f''(x)=96x^2-54x#

#=6(16x^2-9x)#

Now,

#f''(x)=0#

#=>6(16x^2-9x)=0#

#=>x(16x-9)=0#

#=>color(red)(x=0,x=9/(16))# are the inflection points. Inflection points are those points where the curve changes its concavity if any.

graph{x(16x -9) [-5, 5, -5, 5]}

Sign Chart: See image.

Now, to determine the opening of the concavity.

#f''(x)# comes out to be positive. #(+)#.

#f''(x)# comes out to be negative. #(-)#.

#f''(x)# comes out to be positive. #(+)#.

Negative sign indicates that the curve will open downwards. And positive sign indicates it'll open up.

Thus, #(-oo,0)∪(9/(16),oo)# our concavity is upwards.

And,

#(0,9/(16))# our concavity is downwards.

Hope this helps. :)