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?
done giving you freebies. I'll be happy to check your work or point you in the right direction. You must have some idea how to solve such problems.
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.
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.
oops! Thanks for the help
To determine the number of positive integers, "a," such that the function f(x) = x^4 - (76/3)x^3 + 2ax^2 has no local maxima, we need to analyze the behavior of the function.
A local maximum occurs when the slope of the function changes sign from positive to negative at a specific value of x. Therefore, to find the positions of local maxima, we need to determine the critical points of the function.
To find the critical points, we need to take the first derivative of f(x) and set it equal to zero:
f'(x) = 4x^3 - (76/3) * 3x^2 + 4ax = 0
Simplifying the equation:
4x^3 - 76x^2 + 12ax = 0
Now, we need to find the values of "a" for which this equation has no positive solutions. This means that the discriminant of the equation should be negative, i.e., b^2 - 4ac < 0.
Considering ax^2 + bx + c = 0 as a quadratic equation with a = 4, b = -76, and c = 0, we can find the discriminant:
b^2 - 4ac = (-76)^2 - 4 * 4 * 0 = 5776 > 0
Since the discriminant is positive, there are no values of "a" that lead to no positive solutions.
Therefore, there are no positive integers "a" (<1000) for which the function f(x) = x^4 - (76/3)x^3 + 2ax^2 has no local maxima.