# Thread: Continuity - continuous functions

1. ## Continuity - continuous functions

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.

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

(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,

2. 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.

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

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

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

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

3. Originally Posted by Robb
$\displaystyle F(x) = x^3 + 2x+cos^2x$
(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 $\displaystyle F(-1)<0\;\&\;F(0)>0$?
Use the IVT to conclude that there is a root between -1 & 0.

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