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Ulleksa [173]
3 years ago
7

F(x)=-x^(4)-9x^(3)-24x^(2)-16x what is the relative maxium and minimum?

Mathematics
1 answer:
stepladder [879]3 years ago
7 0
We'll need to find the 1st and 2nd derivatives of F(x) to answer that question.

F '(x) = -4x^3 - 27x^2 - 48x - 16     You must set this = to 0 and solve for the 
                                                           roots (which we call "critical values).

F "(x) = -12x^2 - 54x - 48

Now suppose you've found the 3 critical values.  We use the 2nd derivative to determine which of these is associated with a max or min of the function F(x).

Just supposing that 4 were a critical value, we ask whether or not we have a max or min of F(x) there:   

F "(x) = -12x^2 - 54x - 48    becomes    F "(4) = -12(4)^2 - 54(4)
                                                                        =  -192 - 216
                                       Because F "(4) is negative, the graph of the given
                                        function opens down at x=4, and so we have a 
                                        relative max there.  (Remember that "4" is only
                                       an example, and that you must find all three
                                        critical values and then test each one in F "(x).
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Find a factorization of x² + 2x³ + 7x² - 6x + 44, given that<br> −2+i√√7 and 1 - i√/3 are roots.
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A factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

<h3>What are the properties of roots of a polynomial?</h3>
  • The maximum number of roots of a polynomial of degree n is n.
  • For a polynomial with real coefficients, the roots can be real or complex.
  • The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if a+ib is a root, then a-ib is also a root.

If the roots of the polynomial p(x)=ax^4+bx^3+cx^2+dx+e are r_1,r_2,r_3,r_4, then it can be factorized as p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4).

Here, we are to find a factorization of p(x)=x^4+2x^3+7x^2-6x+44. Also, given that -2+i\sqrt{7} and 1-i\sqrt{3} are roots of the polynomial.

Since p(x)=x^4+2x^3+7x^2-6x+44 is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, -2-i\sqrt{7} and 1+i\sqrt{3} are also roots of the given polynomial.

Thus, all the four roots of the polynomial p(x)=x^4+2x^3+7x^2-6x+44, are: r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}.

So, the polynomial p(x)=x^4+2x^3+7x^2-6x+44 can be factorized as follows:

\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)

Therefore, a factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

To know more about factorization, refer: brainly.com/question/25829061

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step by step Explanation

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