How do you verify the intermediate value theorem over the interval [0,5], and find the c that is guaranteed by the theorem such that f(c)=11 where $f(x)=x^2+x-1$ ?

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How do you verify the intermediate value theorem over the interval [0,5], and find the c that is guaranteed by the theorem such that f(c)=11 where $f(x)=x^2+x-1$ ?

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Answer 1 · 2017-09-04T16:39:36

The value of $c=3$

Explanation:

The Intermediate Value Theorem states that if $f(x)$ is a continuous function on the interval $[a,b]$ and $N in (f(a), f(b))$ , then there exists $c in [a,b]$ such that $f(c)=N$

Here,

$f(x)=x^2+x-1$ is continuous on $RR$ as it is a polynomial function.

The interval is $I= [0,5]$

$f(0)=-1$

$f(5)=25+5-1=29$

$f(x) in [ -1, 29]$

Then,

$f(c)=11$ , $=>$ $11 in [-1,29]$

$EE c in [0,5]$ such that $f(c)=11$

Therefore,

$c^2+c-1=11$ , $=>$ , $c^2+c-12=0$

$(c+4)(c-3)=0$

$c=-4$ and $c=3$

$c=3$ and $c in [0,5]$

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How do you verify the intermediate value theorem over the interval [0,5], and find the c that is guaranteed by the theorem such that f(c)=11 where $f(x)=x^2+x-1$ ?

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How do you verify the intermediate value theorem over the interval [0,5], and find the c that is guaranteed by the theorem such that f(c)=11 where f(x)=x^2+x-1f(x)=x^2+x-1?

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How do you verify the intermediate value theorem over the interval [0,5], and find the c that is guaranteed by the theorem such that f(c)=11 where $f(x)=x^2+x-1$ ?