Problem Solving

aflaam messih wrote:if (x-2)^4/4 + (y+2)^4/9 = 4,then the maximum possible value of x is between

a) -1 and 1
b) 1 and 3
c) 3 and 5
d) 5 and 7
e) 7 and 9

I don't know how to solve it. Some help will be highly appreciated.

Given (x-2)^4/4 + (y+2)^4/9 = 4, we can note that since the exponents are even (4) for both (x-2)^4/4 and (y+2)^4/9, they are non-negative numbers . Thus, two non-negative numbers add up 4. To get the maximum value of x, we must have the maximum value of (x-2)^4/4. This follows that we must have the minimum value of (y + 2)^4/4.

Since (y + 2)^4/4 is a non-negative number, the minimum value of (y + 2)^4/4 would be 0.

Again, (x-2)^4/4 + (y+2)^4/9 = 4

=> Maximum possible value of (x-2)^4/4 + Minimum possible value of (y+2)^4/9 = 4

=> Maximum possible value of (x-2)^4/4 + 0 = 4

Maximum possible value of (x-2)^4 = 16

x = 0 or 4

Thus, the maximum possible value of x is 4.

The correct answer: C

Hope this helps!

-Jay
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aflaam messih wrote:if (x-2)^4/4+(y+2)^4/9 =4,then the maximum possible value of x is between

a) -1 and 1
b) 1 and 3
c) 3 and 5
d) 5 and 7
e) 7 and 9

We can maximize the value of x by minimizing the value of (y+2)^4/9. We see that (y+2)^4/9 will be non-negative since (y + 2) is raised to an even power; however, if y = -2, then (y+2)^4/9 = 0. So we have:

(x-2)^4/4 = 4

(x-2)^4 = 16

[(x-2)^4]^(1/4) = ±(16)^(1/4)

x - 2 = ±2

x - 2 = 2

x = 4

x - 2 = -2

x = 0

We see that the maximum value of x is 4, which is between 3 and 5.

Answer: C