Question 224339
Use the Intermediate Value Theorem to show that there is a root of the equation {{{ 2x^3 + x^2 + 2 = 0 }}} in the interval [-2,-1]
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f(x) = 2x^3 + x^2 + 2
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Calculate f(-2):
{{{ 2x^3 + x^2 + 2}}}
{{{ 2(-2)^3 + 2^2 + 2}}}
{{{ 2(-8) + 4 + 2}}}
{{{ -16 + 4 + 2}}}
{{{ -10 }}}
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Calculate f(-1):
{{{ 2x^3 + x^2 + 2}}}
{{{ 2(-1)^3 + (-1)^2 + 2}}}
{{{ 2(-1) + 1 + 2}}}
{{{ -2 + 1 + 2}}}
{{{ 1}}}
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"Mean value theorem" says that if a number (say 0) is between f(-2)=-10 and f(-1)=1 then there will be a value of x that will give you that result.
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Since 0 lies between -10 and 1 there is a value of x.
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To find that value, it is best to use a graphing calculator..
x = -1.197
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Read up more on "intermediate value theorem" at:
http://en.wikipedia.org/wiki/Intermediate_value_theorem