English, asked by shanizeeshan461, 4 months ago

if for a real continuous function fx(), f afbo ()< 0, then in the range of [a b, ] for fx() = 0, there is
(are)​

Answers

Answered by brainlystar1481
0

Answer:

MULTIPLE CHOICE TEST

(All Tests)

BISECTION METHOD

(More on Bisection Method)

NONLINEAR EQUATIONS

(More on Nonlinear Equations)

Pick the most appropriate answer.

Q1. The bisection method of finding roots of nonlinear equations falls under the category of a (an) ______ method.

open

bracketing

graphical

random

Q2. If for a real continuous function f(x), you have f(a)f(b)<0, then in the interval [a,b] for f(x)=0, there is (are)

one root

an undeterminable number of roots

no root

at least one root

Q3. Assuming an initial bracket of [1,5] , the second (at the end of 2 iterations) iterative value of the root of is

0.0

1.5

2.0

3.0

To find the root of f(x)=0, a scientist uses the bisection method. At the beginning of an iteration, the lower and upper guesses of the root are xl and xu, respectively. At the end of this iteration, the absolute relative approximate error in the estimated value of the root would be

For an equation like , a root exists at x=0. The bisection method cannot be adopted to solve this equation in spite of the root existing at x=0 because the function

is a polynomial

has repeated roots at x=0

is always non-negative

has a slope of zero at x=0

The ideal gas law is given by

where where p is the pressure, v is the specific volume, R is the universal gas constant, and T is the absolute temperature. This equation is only accurate for a limited range of pressure and temperature. Vander Waals came up with an equation that was accurate for larger range of pressure and temperature given by

where a and b are empirical constants dependent on a particular gas. Given the value of R=0.08, a=3.592, b=0.04267, p=10 and T=300 (assume all units are consistent), one is going to find the specific volume, v, for the above values. Without finding the solution from the Vander Waals equation, what would be a good initial guess for v?

0

1.2

2.4

4.8

Complete Solution

Multiple choice questions on other topics

Answered by varshika1664
0

Answer:

If for a real continuous function f(x), f(a)f(b) < 0, then in the range of [a,b] for f(x) = 0, there is at least one element in the range.

Explanation:

A continuous function, as its call suggests, is a function whose graph is continuous with none breaks or jumps. i.e., if we're capable of drawing the curve (graph) of a function with out even lifting the pencil, then we are saying that the function is continuous. Studying approximately the continuity of a function is without a doubt vital in calculus as a function can not be differentiable unless it's far continuous. A function is stated to be continuous over an interval if it's far continuous at every and each point on the interval. i.e., over that interval, the graph of the function should not break or jump.

A function f(x) is stated to be a continuous function in calculus at a point x = a if the curve of the function does not break on the point x = a.

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