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7. First I took derivative and got f'(x)= 6x^2-72x+210. divided the whole thing by 6 and found the factors for x^2-12x+35. (x-7)(x-5) which will give me that the smaller critical number is x=5 and the bigger critical number is x=7.

8. First I set the equation to 0 and used the quadratic formula to get that x= -0.94721 and x= -0.05279. Afterwards, I plugged it back into the original equation and got that x=22.90486 which will be absolute maximum and x= 9.541833 which will be the absolute minimum. is this correct?

9. f(x)= x^2 * e^8x , first i took f'(x) and got f'(x)= e^8x * 2x + x^2 * 8e^8x

next , i took the second derivative and got f"(x)= [2x*8e^8x+2e^8x]+[8e^8x*2x + x^2*64e^8x]...

the next step, i have to set it to zero? Can anyone help me with that and tell me if i am doing this correctly please. Thanks a lot.

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# Homework Help: Derivatives, inflections, concavity [included pictures]

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