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strojnjashka [21]
1 year ago
7

What is the maximum number of relative extrema contained in the graph of

Mathematics
1 answer:
maxonik [38]1 year ago
5 0

Step-by-step explanation:

an expression (polynomial) with the highest exponent = n has n zeros (solutions, where y = 0, although some of them can be imaginary numbers, or some can fall together into 1 solution).

and it has max. n-1 relative extrema ("bumps" in the curve, where the curve changes direction from up to down or vice versa).

so, in our case of

f(x) = 3x⁵ - x³ + 4x - 2

we have max. 5-1 = 4 relative extrema.

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Answer:

\huge\boxed{p(x)=4x^3-20x^2+4x+300}

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\text{If}\ x=4\pm3i\ \text{and}\ x=-3\ \text{are the zeros of a polynomial, then it has  a form:}\\\\p(x)=\bigg(x-(4-3i)\bigg)\bigg(x-(4+3i)\bigg)\bigg(x-(-3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x-4+3i)(x-4-3i)(x+3)\bigg(r(x)\bigg)\\\\p(x)=\underbrace{\bigg((x-4)+3i\bigg)\bigg((x-4)-3i\bigg)}_{\text{use}\ (a+b)(a-b)=a^2-b^2}(x+3)\bigg(r(x)\bigg)\\\\p(x)=\bigg((x-4)^2-(3i)^2\bigg)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2

p(x)=(x^2-2(x)(4)+4^2-3^2i^2)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ i^2=-1\\\\p(x)=(x^2-8x+16-9(-1))(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+16+9)(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+25)(x+3)\bigg(r(x)\bigg)\qquad\text{use FOIL}:\ (a+b)(c+d)=ac+ad+bc+bd\\\\p(x)=\bigg((x^2)(x)+(x^2)(3)+(-8x)(x)+(-8x)(3)+(25)(x)+(25)(3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x^3+3x^2-8x^2-24x+25x+75)\bigg(r(x)\bigg)\qquad\text{combine like terms}\\\\p(x)=(x^3-5x^2+x+75)\bigg(r(x)\bigg)

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