Continuity - continuous functions

**Robb** · March 19th 2009, 12:11 AM

Dont quite understand what my lecturer is wanting me to do with this question.. Don't know if i am trying to/thinking its more complicated then it actually is.

$ F(x) = x^3 + 2x+cos^2x $

(a) Explain why there is a number x such that f(x) = 0
(b) Explain why there is only one value of x such that f(x) = 0

Thanks,

**skeeter** · March 19th 2009, 03:01 AM

Originally Posted by Robb

Dont quite understand what my lecturer is wanting me to do with this question.. Don't know if i am trying to/thinking its more complicated then it actually is.

$ F(x) = x^3 + 2x+cos^2x $

(a) Explain why there is a number x such that f(x) = 0
(b) Explain why there is only one value of x such that f(x) = 0

Thanks,

$F(x) = x^3 + 2x + \cos^2{x}$ is continuous.

$F(-1) < 0$ and $F(0) > 0$ ... by the Intermediate Value Theorem, there must exist an $x = c$ , $-1 < c < 0$ , such that $F(c) = 0$ .

$F'(x) = 3x^2 + 2 - 2\cos{x}\sin{x} = 3x^2 + 2 - \sin(2x) > 0$ for all $x$ ... $F(x)$ is always increasing and therefore, can only have a single zero.

**Plato** · March 19th 2009, 03:05 AM

Originally Posted by Robb

$ F(x) = x^3 + 2x+cos^2x$
(a) Explain why there is a number x such that f(x) = 0
(b) Explain why there is only one value of x such that f(x) = 0

Can you see that $F(-1)<0\;\&\;F(0)>0$ ?
Use the IVT to conclude that there is a root between -1 & 0.

Is it true that $F'(x)>0$ ? What does that tell us?

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