How do you verify the intermediate value theorem over the interval [0,3], and find the c that is guaranteed by the theorem such that f(c)=4 where $f (x) = x^{3} - x^{2} + x - 2$ $f(x)=x^3-x^2+x-2$ ?

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How do you verify the intermediate value theorem over the interval [0,3], and find the c that is guaranteed by the theorem such that f(c)=4 where $f (x) = x^{3} - x^{2} + x - 2$ $f(x)=x^3-x^2+x-2$ ?

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Answer 1 · 2017-08-14T19:03:16

$f$ $f$ is a polynomial so $f$ $f$ is continuous on the interval $[0, 3]$ $[0,3]$ .

$f (0) = - 2$ $f(0) = -2$ and $f (3) = 19$ $f(3) = 19$

$4$ $4$ is between $f (0)$ $f(0)$ and $f (3)$ $f(3)$ so IVT tells us that there is a $c$ $c$ in $(0, 3)$ $(0,3)$ with $f (c) = 4$ $f(c) = 4$

Finding the $c$ $c$ requires solving

$x^{3} - x^{2} + x - 2 = 4$ $x^3-x^2+x-2 = 4$ . Which is equivalent to

$x^{3} - x^{2} + x - 6 = 0$ $x^3-x^2+x-6 = 0$

Possible rational zeros are $\pm 1$ $+-1$ , $\pm 2$ $+-2$ , $\pm 3$ $+-3$ , and $\pm 6$ $+-6$ .

Testing shows that $2$ $2$ is a solution.

So $c = 2$ $c = 2$

Answer 2 · 2017-08-14T19:07:09

The value of $c \in (0, 3)$ $c in (0, 3)$ and is $c = 2$ $c=2$

Explanation:

The intermediate value theorem states that if $f (x)$ $f(x)$ is a continuous function on the interval $[a, b]$ $[a,b]$ , then there is a number $p$ $p$ between $f (a)$ $f(a)$ and $f (b)$ $f(b)$ , $(f (a) \neq f (b))$ $(f(a)!=f(b))$ such that there is a number $c \in (a, b)$ $c in (a,b)$ such that $p = f (c)$ $p=f(c)$

Here, $f (x) = x^{3} - x^{2} + x - 2$ $f(x)=x^3-x^2+x-2$ , which is a polynomial function continuous on the Interval $[0, 3]$ $[0,3]$

$f (0) = 0 - 0 + 0 - 2 = - 2$ $f(0)=0-0+0-2=-2$

$f (3) = 3^{3} - 3^{2} + 3 - 2 = 27 - 9 + 3 - 2 = 19$ $f(3)=3^3-3^2+3-2=27-9+3-2=19$

$f (c) \in (f (0, f (3)))$ $f(c) in (f(0, f(3)))$

$f (c) = 4$ $f(c)=4$ such that $c \in (0, 3)$ $c in (0,3)$

Therefore,

$f (c) = c^{3} - c^{2} + c - 2 = 4$ $f(c)=c^3-c^2+c-2=4$

$\Rightarrow$ $=>$ , $c^{3} - c^{2} + c - 6 = 0$ $c^3-c^2+c-6=0$

$f (2) = 2^{3} - 2^{2} + 2 - 2 = 8 - 4 + 2 - 2 = 4$ $f(2)=2^3-2^2+2-2=8-4+2-2=4$

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How do you verify the intermediate value theorem over the interval [0,3], and find the c that is guaranteed by the theorem such that f(c)=4 where $f (x) = x^{3} - x^{2} + x - 2$ $f(x)=x^3-x^2+x-2$ ?

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How do you verify the intermediate value theorem over the interval [0,3], and find the c that is guaranteed by the theorem such that f(c)=4 where f(x)=x^3-x^2+x-2f(x)=x3−x2+x−2f(x)=x^3-x^2+x-2?

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How do you verify the intermediate value theorem over the interval [0,3], and find the c that is guaranteed by the theorem such that f(c)=4 where $f (x) = x^{3} - x^{2} + x - 2$ $f(x)=x^3-x^2+x-2$ ?