$Finding only one real root for a function$

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$Finding only one real root for a function$

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

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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

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$Finding only one real root for a function$

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