Essay Title: 

Algebra 2 Coursework

April 3, 2016 | Author: | Posted in mathematics, mathematics and economics

1-3 . The correct correspondences are 1-b , 2-c , 3-a . In fact , 1 is just the definition of composition of functions 2 is the definition of `f is the inverse of g . We can plot this function (using for example Maple

We evaluate the function at some points : f (0 -4 , f (-2 0 , f (1 9 . Since -2 is a root of the polynomial , we can factor (using the division algorithm for polynomials , see [1] . Therefore f (x ) has three real roots , visible in the next graphic

We can also apply Rolle ‘s theorem on [banner_entry_middle]

the localization of zeros of polynomials . If we compute the values we see that f change its sign at four consecutive points , so f (x ) must have three real roots by Rolle ‘s theorem . The relative maxima and minima must be located between two successive roots so we can estimate : a local maximum near -3 , and a local minimum near -1

Since a polynomial of degree n has at most n roots , we have found all the roots of f (x

The changes of sign tell us that f (x ) has four real roots , a relative minimum near -2 and 3 , a relative maximum near 0 In fact f (-3 0 , f (-2 )0 , f (2 )0 : four changes of sign

Remember : whenever we have a change of sign in an interval , there must be at least one zero in that interval . Moreover , the function must have a relative minimum or maximum between two successive roots . This is a consequence of the more general , and fundamental , Rolle ‘s theorem

So we have also so x 0 is a double solution and x 1 /4 , -1 /4 are the remaining solutions

that is , then we take the square roots . This change of variable allows us to consider our original equation as a quadratic equation in the new variable and then apply the usual formula

implies immediately

4 . To remember (i ) the product of all the roots equals the constant term (up to the sign (ii ) if f (a 0 then x-a divides f (x . So the roots are : 2 , 3 , -4

Note that the factorization is easy to find , by successively adding suitable terms . The rational solutions , if any , must actually be integral divisors of 9 , because the leading coefficient is 1 . So we have to try , 1 ,3 ,9 and their negatives . It turns out that 9 is a root , and we can factor The last factor has no rational roots since its discriminant is not a square

We have used the general theorem : if a polynomial has integral coefficients , then the rational solutions have numerator dividing the constant term and denominator dividing the leading coefficient

so g and f are not inverse to each other



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