If P(x)=4+2sqrt(x+2) and G(x)=4-3X, then what is the largest constant such that P(G(a)) is defined? Thank you!
P(g(x))
= p(4-3x)
= 4 + 2√(4 - 3x + 2)
= 4 + 2√(6-3x)
remember we cannot take the square root of a negative, so
6-3x≥0
-3x ≥ -6
x ≤ 2
so the largest value of a is 2
Well, well, well, if it isn't another math question! Don't worry, I've got my funny hat on and I'm ready to clown around with this one.
To find the largest constant such that P(G(a)) is defined, we need to make sure that the expression inside the square root in P(x) is greater than or equal to zero.
So let's substitute G(a) into P(x) and see what we get:
P(G(a)) = 4 + 2√(G(a) + 2)
Now, we just plug in G(a):
P(G(a)) = 4 + 2√(4 - 3a + 2)
Simplifying further:
P(G(a)) = 4 + 2√(6 - 3a)
For the expression inside the square root to be greater than or equal to zero, we need 6 - 3a to be greater than or equal to zero.
Solving this inequality, we get:
6 - 3a ≥ 0
3a ≤ 6
a ≤ 2
So, the largest constant a such that P(G(a)) is defined is a = 2.
And there you have it, the clown has done the math! Keep on laughing!
To find the largest constant 'a' such that P(G(a)) is defined, we need to determine the domain of the composition of functions P(G(x)).
First, let's substitute G(x) into P(x):
P(G(x)) = 4 + 2√((G(x)) + 2)
Substituting G(a) into P(G(x)):
P(G(a)) = 4 + 2√((G(a)) + 2)
Now, let's find the domain of P(G(a)) by examining the radicand (expression inside the square root).
Since G(x) = 4 - 3x, let's substitute a into G(x):
G(a) = 4 - 3a
Substituting G(a) back into the expression for P(G(a)):
P(G(a)) = 4 + 2√((4 - 3a) + 2) = 4 + 2√(6 - 3a)
The radicand (6 - 3a) must be non-negative, meaning:
6 - 3a ≥ 0
Solving for 'a':
6 ≥ 3a
Dividing both sides by 3:
2 ≥ a
Therefore, the largest constant 'a' such that P(G(a)) is defined is a maximum of 2.
To find the largest constant such that P(G(a)) is defined, we need to determine the values of "a" that make G(a) a valid input for the function P(x).
First, let's find the expression for P(G(a)) by substituting G(a) into P(x):
P(G(a)) = P(4-3a)
Now, we need to ensure that the input 4-3a in P(x) is valid. Looking at the function P(x), notice that the term inside the square root, x + 2, must be greater than or equal to zero to avoid taking the square root of a negative number.
So, we have:
x + 2 ≥ 0
Solving this inequality for x, we get:
x ≥ -2
Now, substitute x with 4-3a to find the valid values for "a":
4-3a ≥ -2
Solving this inequality for "a," we get:
-3a ≥ -6
Dividing both sides of the inequality by -3 (and reversing the inequality sign), we have:
a ≤ 2
Thus, the largest constant such that P(G(a)) is defined is a = 2.