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Interactive video lesson plan for: unit 4 test application of derivatives #3 Points of Inflection second derivative ap calc calculus

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Good day students welcome to mathgotserved.com in this group are going to be going over problem number three, not need for test testing him application of derivatives so we going to be focusing on points of inflection on this tutorial okay so for question number three where to find the points of inflection of the function F of X equals sine X on the interval I'll zero two three pie over to okay so before we get started let's go ahead and go over how to find the points of inflection two X to the point of inflection the x-coordinate of course the point of inflection is something happens it is a point of inflection is F double prime of X change use sign okay changes sign here from positive to negative four from negative to positive so is on the second derivative changes sign of the particle X value then Om that the case that it's a point of inflection right so what were going to do here is fine where the second derivative is set equal to zero that tells us one is neither positive nine negative in the unit does values to create our our boundaries for the intervals where all F double prime is either positive or negative rights is go ahead and find the first derivative first so function is sine X the first derivative F prime of X is equal to cosine X the second derivative of F is all negative sine X okay so on we going to so for where this function is equal to zero W boundary points city solve this equation we have sine X equals zero the left times by negative and when Islam sine X equal to zero so on the solution to this equation when is sine X equals zero we notice sine X is equal to zero whenever how X is an integer multiple of pie Om so in this restricted interval right here the solution is going to be on zero and five okay so our solution the general solution to this equation is X is equal to and prime where and is an integer okay another way to write it says the can save X is equal to zero plus or minus five plus or minus two pie in a pattern goes on and on okay but or you can clearly see that can zero two three over to pie one half I only zero and high all our solution so case of X is equal to zero and I in the intervals on the interval I'll zero, three pie over to okay so just to give you visual this is the graph of the second derivative so the second derivative we have the going from zero two three five two so let's call this our Om over to was five and then right here three five or two okay so graphing negative sine our with the graph of the sign the same thing as the second derivative just to be precise so that the negative sine like to negative sine starting from the center to the main back to the center and then to the maximum okay so this right here the graph of the second derivative to sketch okay so this is F double prime of X now what it is equal to zero you can clearly see that the second derivative is equal to zero at zero and pie okay now Om we looking for is concave up and concave down so what we looking for is the sign of the second derivative right so we going from zero to pie and then three pie over to what was happening here is one of the endpoints happens to be of points where is to the second derivative is zero also okay so have an overlap this endpoint right here okay right so what the sign of F double prime on interval I'll why and to one right to that now this problem is the have a visual here we have the graph of the second derivative so can easily tell that zero and pie the second derivative negative in between pie and three five two the second derivative is positive I so those of the signs of F double prime so sine of F double prime what is the second derivative tell us about the function itself it tells is where is concave up or concave down from zero to pie second derivative is negative which means the function is concave down on this interval and concave up from pie to three pie over to okay so is go ahead and write it down the function the function is all actually were not asked to find where is concave up or concave down we have to find points of inflection okay so that the state this Om so let's go ahead and state the points of inflection our sin concavity changes am I and that's a point of inflection okay so we can at the concavity what the sign of the second derivative let's talk about use the sign of the second derivative right so since F double prime change use from positive on the from negative to positive from negative to positive on time from negative to positive using negative to positive or from concave down to concave up about a concavity to the second about as talked about the sign of the second derivative so since F double prime changes from negative to positive at X equals pie there is equal to of inflection inflection at that point at X equals pie right so we looking for a point of inflection

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