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Show that the curves $y=e^{-x}$ and $y=-e^{-x}$ touch the curve $y=e^{-x} \sin x$ at inflection points.

$=-e^{\pi / 2}$

Calculus 1 / AB

Chapter 4

APPLICATIONS OF DIFFERENTIATION

Section 3

Derivatives and the Shapes of Graphs

Derivatives

Differentiation

Applications of the Derivative

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Campbell University

Idaho State University

Lectures

04:40

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.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

03:40

Show that the curves $ y =…

01:08

Show that the curve $y=a x…

03:55

Show that the inflection p…

0:00

02:44

02:39

Show that the function $f(…

for this program. We're looking for the inflection points so we can write out a second security first. So this is a perfect second curative off wife. So the impression appointment satisfy white after Prime Mikel zero because the exponential function is known Next office unknown zero. So I reduced this equation reduced to co sign X equals zero. That means X equals due to K pass one over. I took her plus one times pi over two for Keiko's zero past minus one past minus two. So have infinitely many inflection point there. Um, so if x equals to this value And, uh, so it's, um, right out in this way. So if x equals two, um, two que pass one tie over to the corresponding y value we will be you actually minus, um two Que Pass one high over two times. Sign off. Two que past one. Hi. Over to. So this is the point. Uh, this is the inflection point. So But look at this. This this part Now, if we graph sigh on the coordinates for each two K plus one high over to, um sign off Took a pass. One powerful to issue. Um one or minus one. So all the inflection point has this form to okay, plus one high over to and ah, uh, minus one to the two. OK, you to the minus two k past one terms pilot to so off this inflection point are either on whyy closed toe E to the minus X here, WHYY close to minus C to remind the sex. Then we finish the proof. So, uh, both both curve e to the minus X and the minus eat reminding sex were touched, uh, dysfunction at its inflection points because off its inflection points has have this this kind of form and that this phone satisfy the following to our functions.

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