Answer:
10.29 u
Step-by-step explanation:
<u>Given :- </u>
And we need to find out the distance between the two points . So , here we can use the distance formula to find out the distance. As,
D = √{(x2-x1)² + (y2-y1)²}
D =√[ (7+2)² +(9-4)²]
D =√[ 9² +5²]
D =√[ 81 +25]
D = 10.29
Answer:
Step-by-step explanation:
I found these using a graphing calculator. (See attached)
The function is a polynomial of 4th degree with a negative leading coefficient. (The coefficient of the 4th-degree term is -3.) Therefore, it opens downward and has a range that extends to -∞. This function only has one global maximum and no local minimum.
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The maximum can be found by solving the cubic equation that results from taking the derivative of this function and finding the root of that:
df/dx = -12x^3 -2x -20 = 0
There are formulas for the solutions of a cubic, but the one negative real root of this cubic can be found easily with a graphing calculator using iteration:
x ≈ -1.13879908894.
Dividing by -2, the above derivative function can be written as ...
g(x) = 6x^3 +x +10
The one real zero of it can be found using Newton's Method iteration.
The Newton's Method iterator function is
h(x) = x - g(x)/g'(x)
where g'(x) is the derivative of g(x), often available as a function on the graphing calculator.
The starting value for the iteration can be the x-value obtained from the graph: -1.139. It only takes about 2 or 3 iterations to find x to the precision of the calculator: x = h(h(x)). This x is the x-coordinate of the maximum, so f(x) is the maximum value: 29.4335546516.
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The attachments show the function definitions and their use for finding the maximum of f(x) using a TI-84 calculator. Y₁ is the function f(x); Y₂ is its derivative, shown above as g(x). Y₃ is the iterating function, shown above as h(x). The nDeriv( ) function is a calculator function that finds the numerical derivative of its argument.
