Let f(x) = 3x + 1 and g(x) = 2 - x. Determine the values for a such that

A) (a)
B) f(a^2) = g(2a)

pog

A) To find the values for a such that (a), we need to equate the two functions and solve for a:

(a)
3a + 1 = 2 - a

Now let's solve for a:

3a + a = 2 - 1
4a = 1
a = 1/4

So, the value for a that satisfies (a) is a = 1/4.

B) To find the values for a such that f(a^2) = g(2a), we need to substitute the expressions f(a^2) and g(2a) into the equation and solve for a:

f(a^2) = g(2a)
3(a^2) + 1 = 2 - (2a)

Now let's solve for a:

3a^2 + 2a - 1 = 0

Hmm, it seems that we have a quadratic equation here. Let's try to factor it:

(3a - 1)(a + 1) = 0

Now, using zero-product property, we can set each factor equal to zero:

3a - 1 = 0 or a + 1 = 0

Solving each equation:

3a = 1 or a = -1
a = 1/3

So, the values for a that satisfy f(a^2) = g(2a) are a = 1/3 and a = -1.

Please remember to take these answers with a grain of salt. I'm just a clown bot, after all!

To determine the values for a that satisfy the given conditions, we need to substitute the expressions for f(x) and g(x) into the equations and solve for a.

A) (a)
Substituting f(x) = 3x + 1 and g(x) = 2 - x:
3a + 1 = 2 - a

To solve for a, we can combine like terms:
3a + a = 2 - 1
4a = 1

Dividing both sides by 4, we find:
a = 1/4

Therefore, the value of a that satisfies (a) is a = 1/4.

B) f(a^2) = g(2a)
Substituting f(x) = 3x + 1 and g(x) = 2 - x:
3(a^2) + 1 = 2 - (2a)

To solve for a, we need to simplify the equation:
3a^2 + 1 = 2 - 2a

Combine like terms:
3a^2 + 2a + 1 = 2

Rearrange the equation:
3a^2 + 2a - 1 = 0

To solve this quadratic equation, you can use the quadratic formula:
a = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 3, b = 2, and c = -1. Substituting the values into the formula:

a = (-2 ± √(2^2 - 4(3)(-1))) / (2(3))
a = (-2 ± √(4 + 12)) / 6
a = (-2 ± √16) / 6
a = (-2 ± 4) / 6

This gives us two possible solutions for a:
a1 = (4 - 2) / 6 = 2/6 = 1/3
a2 = (-2 - 4) / 6 = -6/6 = -1

Therefore, the values of a that satisfy f(a^2) = g(2a) are a = 1/3 and a = -1.

Sub "a" in where x is...

A) 3a+1 = 2 -a and then solve for a
B) where ever you see an "x" in the first function set it to be a^2, then in g(x) where you see the x let it be (2a) and then solve for a
So you have 3a^2 + 1 = 2 - 2a move everything to the right hand side and factor : )

Thank you Ms Pi_3.14159265358979

You are very welcome! It is great to see diligent students working on math : )