Solve the following algebra + calculus question on functions. .
Maximum , minimum values over domains.
Answers
Min max value. .
The given question is probably wrong...
Perhaps it's f(x) = sqrt[a - sqrt(a+x)].
g(x) = [a + sqrt(a - x)].
To find out the value of a for
A) min of f(x) = min g(x).
B) min f(x) < min g(x).
C) min f(x) > min g(x).
You can find 1st differentials. They are always negative. Or, you could see directly that as x increases, both functions decrease. So both functions are minimum at the right end of their domains.
f(x) is defined if sqrt(a+x) is defined. ie.. x > -a.
Also f(x) defined if iff a >= sqrt(a+x)
ie. x <= a^2 -a.
So f(x) min is at x= a^2 -a. ie. It's 0.
That f(x) min is zero could be inferred directly from given f(x) as it is a square root and non negative.
Similarly, derive that g(x) is defined for x in (-infty, a].
Also note that a > 0 for f(x) to be defined. Else f(x) will be imaginary.
Now as g(x) is always decreasing, it's minimum when x = a at right end of domain.
Min g(x) = sqrt(a).
NOW:
A) 0 = sqrt(a)
=> a = 0. NOT possible.
B) 0 > sqrt(a)
=> Not possible.
C) 0 < sqrt (a)
=> always true for all x over which f and g are both defined.
This much analysis isn't needed to solve it.. But explained so that it's logic is understood better.
*==> ALTERNATELY ==*
Correct question is probably :
f(x) = sqrt(a - sqrt(a+x)] - x or + x or -a or +a etc.
...
g(x) = sqrt[a + sqrt(a-x)] + x or -x or -a or +a etc...
Answer:
hope this answer useful to you mate I
