do 4 only
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
thats easy , first find where derivative is zero
so you know where the function turns around ,
then compare the values of the function at the turning points
Not the answer you are looking for? Search for more explanations.
if you go from positive to negative , or negative to positive function value between two turning points then you must have an intersection with the x axis ( ie a zero )
if the sign of the function value doesnt change then there are no roots in the interval .
*no real roots
find the number of the real roots , then to find the number of complex root you subtract the number of real roots from the degree of the function
Or you can use Descartes theorem and count the sign changes in f(x) for positive roots and f(-x) for neg roots
positive: 2 or 4
imaginary: 0 or 2
By the fundamental theorem of algebra, this polynomial will have five roots.
Descartes rule of signs has that r has either two or zero positive real roots, and one negative real root.
The rational root theorem says that 1 is a root of r, and since this is real and positive, r clearly does not have zero real positive roots - it must have two.
So, we have so far:
Two real positive roots
One real negative root
Since r has rational coefficients, the remaining roots are complex conjugates by the complex conjugate root theorem.
So for r:
Two real, positive roots
One real, negative root
Two complex conjugate roots