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unit 4 test application of derivatives #2 Mean value theorem

838 views · Published 8 December 2014 · 8:16 · Indexed 20 September 2026

Channel: maths gotserved · 2014 · Education

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Good day students welcome to mathgotserved.com in this clip were going to be going over how to apply the mean value there finding the C that satisfies its conclusion this is number two of our unit for tests on the application of derivatives okay so problem two reads given that F of X is equal to two Exodus third minus four X plus three find the numbers see in negative one two that satisfy the mean value there for this function so the assumption is that on the initial conditions are already met with can clearly see that this function is differentiable formula numbers and is also continuous so we know that the initial conditions are already met so we looking for the sine foresees that satisfy the conclusion are so lovely value there is the view that is as follows MVT so is it F some function F okay if F is continuous on the closed interval the be and differentiable on the open interval a be there exists a C F prime of C the slope of the tangent line at X equals C is equal to the average rate of change from eight to be the limit of a divided by be minus eight right okay let's go ahead and do this we going to bring this up into two parts Om on will do the derivative part in blue of the agreement F prime of C and then alter the other part okay on that in black this is the instantaneous rate of change at X equals C and in the average rate of change will do that part in black okay right now let's deal with this site first let's find F prime of C so how do we do that is recall what the function is F of X is to X to the third minus four X plus three F prime of C simply means that violates the derivative of this function at X equals the find a derivative plugging C for X and you are done okay so we have F prime of X the derivative term right term differentiation we going to have six X square minus four derivative of three zero so the whole that F prime of C is equal to six C square minus four so the instantaneous rate of change for the slope of the tangent line at X equals the is six C square minus four okay now let's go ahead and find the average rate of change that the up and we going to second equal to each other right so here we need to know be a F of be in F of a so we have it has about negative one two so this is a and this is the right so eighty is negative one the is to F of a is F of negative one so I'm going to plug negative one into the original function of the derivative so be careful with that so have two times negative one to the third minus four times negative one plus three if you come to that's using your order of operations you end up with eleven as a value of F of be F of be related about weight the function this original function at X equals two so we going to have two times two to the third minus four times two plus three compute that end up with five right now let's go ahead and find the average rate of change of the slope of the secant line that they to the F of five F of two minus F of negative one over all two minus negative one okay right compute at what F of two F of negative one are so we're going to have on sorry X which surrounds Om F of two is eleven and F negative one is five right so this plug it into this formula right here we going to have see F of two is eleven minus F of negative one on divided by two minus minus one is negative three all eleven minus five six over three which is to this is the average rate of change this is the instantaneous rate of change within a set them equal to each other so going to be solving for to see where six C square minus four equals the average of change to five Stratus all that Om you divide these at sports = X six C square equals six divided by six C square equals one we take the square roots you going to have see equals plus or minus the square root of one which is plus four minus one okay before we identify our see that satisfy the conclusion let's ensure that these two values are in the interval I'll negative one, to so if you notice negative one is that endpoints is must include must be excluded okay so only positive one is in the interval I'll negative one, to okay the other value is endpoint and and endpoint is not included foresees that satisfies the conclusion of the MVT alright so let's write down our final statements the C that satisfies the conclusion of the mean value theorem Stratus MVT and is see equals positive one right to that's that thanks so much for taking the time to watch this presentation really appreciated feel free to subscribe to our channel for updates to other cool tutorials such as this you have any questions or comments to free to included in the comments section markets,, mathgotserved.com thanks again for watching and have a wonderful day

unit 4 test application of derivatives #2 Mean value theorem

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