Find the polynomial of degree 6 which satisfies limx →0(1+(J(x))/(x^3))^x=e^2 and has local maxi

Find the polynomial of degree 6 which satisfies limx...

Key Concepts

Taylor Series and Asymptotic Analysis

Expanding a function in a Taylor series about a specific point allows one to approximate its behavior in the neighborhood of that point. This is particularly useful for evaluating limits or matching asymptotic behavior by comparing coefficients, as it enables the accurate representation of small?x behavior in functions that involve both polynomial and exponential terms.

Polynomial Function Construction

This concept involves building a polynomial that meets a set of predetermined conditions. In many problems, such as when specific behavior or values at certain points are required, one expresses the polynomial in its general form with unknown coefficients and then determines these coefficients using the given constraints.

Degree and Coefficient Determination

A polynomial of a fixed degree contains a predetermined number of coefficients. These coefficients can be found by setting up a system of equations using provided conditions—for example, by prescribing the polynomial's value and the values of its derivatives at various points. This method ensures that the polynomial fits exactly to the desired behavior.

Exponential Limit and Logarithmic Transformation

When limits involve expressions of the form (1 + f(x))^(g(x)) and are set equal to an exponential, a common technique is to take the logarithm of the expression. This transforms the problem into finding the limit of a product (often x multiplied by a logarithmic term) which can then be handled by standard limit or Taylor expansion techniques.

Local Extrema and Derivative Tests

Local maximum and minimum points of a function occur where its derivative is zero (the critical points). The first derivative test identifies these potential points, while the second derivative (or other higher derivative) tests are then used to classify each critical point as a local maximum or minimum. This process is key in ensuring that the function exhibits the prescribed extremal behavior.

Transcript

00:01 In this question, we need to find a polynomial of degree 6 with satisfied the given condition.

00:07 The given condition is limit extending to 0, 1 plus fx divided by x q to the power x is equal to e square.

00:23 Okay, it should satisfy the given condition plus it contains its local maxima at x equals to 1.

00:31 And it contains its local minima at x equals to 0 and x equals to 2 okay these are the three conditions which are given to find out this function first of all note that this function fx cannot have degree less than 3 okay because if contains if it contains degree less than 3 then this limit will be undefined okay so first of all we can say that effects will contain a degree always greater than 3 okay so let's say fx equals to a x to the power 4 plus b x to the power 5 plus c x to the power 6 okay because it is saying that it is a polynomial of degree 6 okay now note that if we put the value of x in the given condition it becomes limit extending to 0 1 plus a x because i have divided by x cube a x plus bx square plus c x cube to the power 1 by x is equals to e square now note that to solve this kind of inequality what we have to do we always have to take the logg both side because here we can see the power 1 by x right so what we'll do is we will take log both side then it becomes limit to 0 it becomes log 1 plus x plus px square plus cx cube into 1 by x okay 1 by x will come in the front and it becomes log e square and log e square is nothing but 2 okay now if we use the expansion of log 1 plus x okay remember the expression the expansion of log 1 plus x okay remember the expression the expansion of log 1 plus x is nothing but x minus x square by 2 plus x cube by 3 and so on so now we'll just come back to our expression it becomes limit extending to 0 a x plus bx square plus c x cubed divided by x is equals to 2 okay because we'll take log 1 plus x just as x and we'll neglect the higher terms so it just becomes this now we can clearly see that if we separate these terms then these two terms will become zero okay these two term will not have anything else because it contains x tending to zero so we can say that a becomes two okay so from here we arrive at a condition that a equals to two now from the other two conditions which are given, we will find b and c.

03:46 Okay, now we know that fx is equals to 2x to the power 4 plus bx to the power 5 plus c x to the power 6.

03:58 Okay, so we can find out f dash x.

04:01 F dash x comes out to be 8x cube plus 5b x x to the x to the part plus c c.

04:12 6c x to the power 5 okay now we are given that this function contains its maximum value at 1 okay so if this function contains its maximum value at 1 then and it contains its minimum value at 0 and 2 okay so the f -dice 1 will be 0 similarly f -2 will be 0 similarly f -0 -0 and maximum values of the function.

04:45 Now we'll not have this condition because from this condition we arrive and nothing, okay, because this function literally becomes zero.

04:52 So we'll look at f -1...

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