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A) dec on [-4,2], inc [2,8]
B) inc on [-4,4], dec [4, infinity)
C) inc on [-4,2] dec on [2,8]
D) inc on [-4,4] dec on [2,infinity)
E) inc on (-inf, 2] dec on [2,8]

2007-03-25 21:36:57 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

Hi,

The correct answer is C. If you graph y1 = (8 - x)^.5(x + 4)^.5 on your calculator, you will see the graph of this function. It looks something like the top of an oval. Note that the expressions raised to the .5 or 1/2 power are really square roots, so in order to be able to take the square root of those expressions (8 - x ) limits x to numbers 8 or smaller while (x + 4) limits x to numbers -4 or larger. If you use the MAX command on the TI-83 calculator under CALC, you will see that the MAX value occurs at (2,6). Since the smallest x value possible was -4 and the max was at x = 2, the function is increasing from [-4, 2]. Then from x = 2 to x = 8 the graph is getting lower , so it is decreasing from [-2,-8].

I hope this helps!

2007-03-25 21:54:06 · answer #1 · answered by Pi R Squared 7 · 0 0

In order to find where the function is increasing or decreasing, we simply take the derivative of the function, which is
f'(x)=(4-2x)/(2(-x^2+4x+32))^(1/2)
Then we substitute in certain numbers to determine the sign of the slope. A negative number will indicate a decreasing section, and a positive number will indicate an increasing section.

The function f is not defined for values of x less than -4 or greater than 8, since sqrt(negative)=imaginary. So that only leaves the two possibilities, a and c. To see if the function is increasing or decreasing on [-4,2], we need only to plug in a value for x into f'(x) for some x in the interval. I arbitrarily choose 0, which gives 2/sqrt(32), which is positive, so the function f is increasing on this interval. Thus the answer is

C) inc on [-4,2] dec on [2,8]

2007-03-26 05:01:43 · answer #2 · answered by jnamnath 1 · 0 0

first you can put all under the square root

you have ((8-x) (x+4))^1/2

The function is defined only between -4 and +8, else square root of negative number

can also be written as (-x^2+4x+32)^0.5

dy/dx = 0.5(-x^2+4x+32)^(-0.5)(-2x+4)
dy/dx =0 for x=2 , positive under x=2, negative over
so from -infinity to -4 not defined
from -4 to +2 increasing
from +2 to +8 decreasing
Over +8 not defined

2007-03-26 05:00:22 · answer #3 · answered by maussy 7 · 0 0

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