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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) $
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03:09
Daniel Jaimes
Calculus 1 / AB
Chapter 2
Limits and Derivatives
Section 5
Continuity
Limits
Derivatives
Campbell University
University of Michigan - Ann Arbor
University of Nottingham
Boston College
Lectures
04:40
In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.
In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.
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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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