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Prove, without graphing, that the graph of the function has at least two $ x $-intercepts in the specified interval.

$ y = \sin x^3 $, $ (1, 2) $

Let $f(x)=\sin x^{3} .$ Then $f$ is continuous on [1,2] since $f$ is the composite of the sine function and the cubing function, bothof which are continuous on $\mathbb{R}$. The zeros of the sine are at $n \pi$, so we note that $0<1<\pi<\frac{3}{2} \pi<2 \pi<8<3 \pi$, and that thepertinent cube roots are related by $1<\sqrt[3]{\frac{3}{2} \pi}[\text { call this value } A]<2 .[\text { By observation, we might notice that } x=\sqrt[3]{\pi}$ and$x=\sqrt[3]{2 \pi} \text { are zeros of } f .]$Now $f(1)=\sin 1>0, f(A)=\sin \frac{3}{2} \pi=-1<0,$ and $f(2)=\sin 8>0 .$ Applying the Intermediate Value Theorem on$[1, A]$ and then on $[A, 2],$ we see there are numbers $c$ and $d$ in $(1, A)$ and $(A, 2)$ such that $f(c)=f(d)=0 .$ Thus, $f$ has atleast two $x$ -intercepts in (1,2)

03:09

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

Campbell University

Oregon State University

Baylor University

University of Nottingham

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for this question. We're going to use the intermediate value theorem to prove the question. So the I V T states that if ffx is continuous on interval A B such that F F a is sequel is less than zero and ff b is greater than zero or vice versa. Um There exists an X not belonging to the interval A. B. Such that F of X zero is equal to zero. So our interval that's given here is 1:02 and our function is sign up, execute so let us break this inter uh inter roll down into one comma 1.5 And 1.5 coma too to see the behavior at each of them. So considering the interval one comma 1.5 um RFs one is equal to Sign of one. Cute. Yeah, which is zero point 84 and F of 1.5 would be a sign of 1.5 Cubed, which is negative. Zero point 23, notice how are f of one is positive and are F f At 1.5 is negative. So applying the IBT here, there has to be at least one point where F of x not is equal to zero, so one X intercept in this interval at least when I intercept. Um And considering the other interval, 1.5, RF for 1.5 has found previously As negative 0.23 and f of two is a sign of eight which is 0.99 and again notice how are F 1.5 is negative Mueller F of two is positive. So applying the IBT here again, there has to be at least one point in this interval Where f effects not as equal to zero. So there has to be at least one X intercept. Oops X intercept in this interval. So, um combining these two intervals, we need to have at least two X intercept because we had one, at least one X intercept in the interval one comma 1.5 And at least one intercept in 1.5 comma two.

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