Asked by andy
How many positive integers a (<1000) are there such that the function
f(x)=x^4−76/3x^3+2ax^2
has no local maxima?
f(x)=x^4−76/3x^3+2ax^2
has no local maxima?
Answers
Answered by
Steve
done giving you freebies. I'll be happy to check your work or point you in the right direction. You must have <b>some</b> idea how to solve such problems.
Answered by
andy
alright, so i got that
f'(x)=4x^3-76x^2+4ax
=4x(x^2-19x+4a)
since this must never equal 0, it must have no solutions. Therefore, x must not be 0 and x^2-19x+4a must not be 0. For that to occur, 289-16a<0, so a > 18.0625.
Can you please check me work?
Sorry for bothering you so much.
f'(x)=4x^3-76x^2+4ax
=4x(x^2-19x+4a)
since this must never equal 0, it must have no solutions. Therefore, x must not be 0 and x^2-19x+4a must not be 0. For that to occur, 289-16a<0, so a > 18.0625.
Can you please check me work?
Sorry for bothering you so much.
Answered by
Steve
watch the algebra!
f'(x)=4x^3-76x^2+4ax
= 4x(x^2-19x+a)
The discriminant 361-4a must be negative, so a > 361/4
You could have seen that your answer was incorrect by graphing the resultant polynomial.
f'(x)=4x^3-76x^2+4ax
= 4x(x^2-19x+a)
The discriminant 361-4a must be negative, so a > 361/4
You could have seen that your answer was incorrect by graphing the resultant polynomial.
Answered by
andy
oops! Thanks for the help
There are no AI answers yet. The ability to request AI answers is coming soon!