Question #d2272

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Question #d2272

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Answer 1 · 2017-11-09T17:04:33

The interval is $= (- 0.875, - 0.8671875)$ $=(-0.875,-0.8671875)$

Explanation:

The Intermediate Value Theorem States that :

If $f (x)$ $f(x)$ is a continuos function on the interval $[a, b]$ $[a,b]$ and $N$ $N$ is any number between $f (a)$ $f(a)$ and $f (b)$ $f(b)$ where $f (a) \neq f (b)$ $f(a)!=f(b)$ , then there exists a number $c$ $c$ in the interval $(a, b)$ $(a,b)$ such that $f (c) = N$ $f(c)=N$

Here,

$f (x) = x^{5} - x^{2} + 2 x + 3$ $f(x)=x^5-x^2+2x+3$

$f (0) = 0 - 0 + 0 + 3 = 3$ $f(0)=0-0+0+3=3$

$f (- 1) = {(- 1)}^{5} - {(- 1)}^{2} + 2 (- 1) + 3 = - 1$ $f(-1)=(-1)^5-(-1)^2+2(-1)+3=-1$

$\exists c \in (- 1, 0)$ $EE c in (-1,0)$ such that $f (c) = 0$ $f(c)=0$

$f (- 0.5) = 1.71875$ $f(-0.5)=1.71875$

So, $c \in (- 1, - 0.5)$ $c in (-1,-0.5)$

$f (- 0.75) = 0.7$ $f(-0.75)=0.7$

$c \in (- 1, - 0.75)$ $c in (-1,-0.75)$

$f (- 0.875) = - 0.0285$ $f(-0.875)=-0.0285$

$c \in (- 0.875, - 0.75)$ $c in (-0.875,-0.75)$

$f (- 0.8125) = 0.361$ $f(-0.8125)=0.361$

$c \in (- 0.875, - 0.8125)$ $c in (-0.875,-0.8125)$

$f (- 0.84375) = 0.173$ $f(-0.84375)=0.173$

$c \in (- 0.875, - 0.84375)$ $c in (-0.875,-0.84375)$

#f(-0.859375)=0.074

Therefore,

$x \in (- 0.875, - 0.859375)$ $x in (-0.875,-0.859375)$

$f (- 0.8671875) = 0.023$ $f(-0.8671875)=0.023$

So,

$c \in (- 0.875, - 0.8671875)$ $c in(-0.875,-0.8671875)$

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Question #d2272

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