Determine whether the following can be modeled by a quadratic,
1 answer:
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Answer:
A. x = 11/16
Step-by-step explanation:
For the purpose here, it is convenient to rearrange the equation to f(x) -g(x) = 0. We know the root will be in the interval [0, 1] because (f-g)(0) = -3 and (f-g)(1) = +3. At each iteration, we evaluate (f-g)(x) at the midpoint of the interval to see which of the interval end points can be moved and still bracket the root.
Using the bisection method starting with the interval [0, 1] we find f(1/2)-g(1/2) < 0, so we can move the interval limits to [1/2, 1].
For the next iteration, we find f(3/4) -g(3/4) > 0, so we can move the interval limits to [1/2, 3/4].
For the third iteration, we find f(5/8) -g(5/8) < 0, so we can move the interval limits to [5/8, 3/4].
Then the root is approximately the middle of that interval:
x ≈ (5/8 +3/4)/2 = 11/16
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This value of x is 0.6875. The root is closer to 0.639802004233. The bisection method takes about 3 iterations for each decimal place of accuracy. Other methods can nearly double the number of accurate decimal places on each iteration.
Answer: The correct option is second, i.e., x>0.
Explanation:
As we know that the domain is the set of all possible inputs. If function is defined as, f(x) then all possible value of x for which the function f(x) is defined is called domain.
In a graph the domain is defined on the x axis and the range of the function is defined on y-axis.
In the given graph the function is defined from x=0 to because for
the graph is not defined. It means for
the function is not defined.
Sicen the graph is defined for all positive values of x, therefore the domain of the function is al real number greater than 0. It can be written as x>0 and the second option is correct.