Each polynomial function has exactly one positive zero. Use the method of Example 10 to approxim

Each polynomial function has exactly one positive zero. Use...

Key Concepts

Numerical Root-Finding Methods (Bisection Method)

Numerical root-finding methods, such as the bisection method, are iterative techniques used to approximate the zeros of a function when exact algebraic methods are infeasible. The bisection method, in particular, repeatedly narrows an interval where the function changes sign by halving it, thereby converging to a zero with a desired level of accuracy. This method is based on the IVT and is reliable for continuous functions.

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a principle in calculus which states that if a continuous function takes on values of opposite sign at two points, then there exists at least one root in the interval between those points. This theorem provides the theoretical foundation for many numerical methods used to approximate zeros, ensuring that a zero actually lies within a selected interval.

Zeros of Functions

The zeros of a function are the points where the function evaluates to zero. For polynomial functions, finding the zeros is essential in understanding the function's behavior. In many cases, particularly with higher degree polynomials, algebraic solutions may be difficult or impossible, hence the need for numerical approximation methods.

Polynomial Functions

Polynomial functions are expressions involving sums of powers of a variable with constant coefficients. They are continuous, smooth, and differentiable everywhere, which means their graphs do not have any breaks or jumps. Understanding the behavior of polynomial functions, especially their end behavior and turning points, is fundamental when analyzing where the function might cross the x-axis (i.e., where it has zeros).

Transcript

00:01 The given f of x that is equals to 2x to the power 4 minus 3x q minus 8 x squared sorry it's minus 4x square minus 4x square minus 8 and with the help of that we need to find exactly 1 positive 0 at which point this function has exactly 1 positive 0 and we have to find the value up to 5.

00:31 Two decimal places only.

00:33 So first of all what we are going to do first of all we will find that among which two integers i mean like between which two integers the value of zero that will be lying okay so first of all we'll check at f of one so at f of one it would like two minus three minus four minus eight so obviously it will like two minus three that will be minus one minus one minus four minus five minus eight that will be minus of 13 now let's check for f of 2 so for f of 2 it will like 2 multiplied with 2 to 2 to the power 4 that will be 16 so 2 multiplied with 16 minus 3 multiplied with 2 cube so that will be equal to 8 3 multiplied with 8 minus 4 multiplied with 2 square that will be equal to 4 minus of 8 so on solving it will be equal to 32 minus 24 minus 16 minus 8 so it will like 32 minus 24 that will be 8 minus 8 that will cancel out each other so it's like minus of 16 so obviously it is confirmed that between 1 and 2 the so now let's check for 3 so it's like our f of 3 that will be equal to 2 multiplied with 3 to the power 4 so that would be 81 minus 3 multiplied with 3 2 so that will be sorry it's 3 2 so let me make it clear 3 2 is nothing but 27 so we'll have over 27 over here minus 4 multiplied with 3 square so that will be 9 minus 8 so on solving it will like 2 multiplied with 81 it will be 162 minus 81 minus 36 minus 8 so it will be 81 minus 44 which will obviously be positive number so it's confirmed that between 2 and 3 the 0 will be lying now we'll check between 2 and 3 the 0 will be lying.

02:32 So it's like f of 2 on putting value it will minus of 60 for f of 2 .1 on putting the value it will be equals to minus of 14 .52688.

02:50 Similarly for f of 2 .2 on putting the value will have minus 12 .4528.

03:00 Similarly for f of 2 .3 it will be equals to minus 9 .6928 again f of 2 .4 so that will be equals to minus 6 .1568 now for f of 2 .5 for f of 2 .5 it will be like minus 1 .75 now for f of 2 .6 it will be equal to 3 .6272 so now if you look carefully the sign is changing over here.

03:45 It means that our 0 will be lying somewhere between f of 2 .5 and f of 2 .6.

03:50 So now we'll check between f of 2 .5 and f of 2 .6...

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