$Finding only one real root for a function$

Question

$Finding only one real root for a function$

Let $h(x)=\frac{9}{2} x-4sin(x)$, Show that h(x) has exactly one real root.

My professor also gave a little "hint"
(This one real root should be really easy to find. In fact you shouldn't have to "solve" anything to find it. All you have to show is there cannot be another one.)

Would I use the Intermediate Value Theorem? I'm not really sure where to start :(

I already found h'(x) which is $\frac{9}{2} -4cos(x)$
I also tried to set h(x) equal to 0 and got $\frac{8}{9} sin(x)$, I'm not sure if any of those are helpful or not. I'm probably overthinking this entire problem LOL

Thanks in advance

Calculus

Ricekrispiefish

6

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

View the answer

Let $h(x)=\frac{9}{2} x-4 \sin(x)$. Then
\[h'(x)=\frac{9}{2} -4 \cos x \geq \frac{9}{2} -4 \times 1 =\frac{1}{2}. \]

Hence $h'(x)\geq \frac{1}{2}>0$. Hence $h$ is a strictly increasing function, and any strictly increasing functions can have at most one solution.

Method #2. Suppose $h$ has two roots $x_1, x_2$. Then by mean value theorem there exists $c$ such that
\[\frac{1}{2}\leq h'(c)=\frac{h(x_2)-h(x_1)}{x_2 -x_1}=\frac{0-0}{x_2 -x_1}=0.\]
Hence $\frac{1}{2} \leq 0$ (!) which is a contradiction. So $h$ can not have more than one root.

Erdos

4.8K

$Finding only one real root for a function$

Answer

Related Questions

Search