Please help! Need answered! Please and thank you.
1 answer:
Answer:
1. lim x→0 7e^x =7
2. lim f(x) does not exist
f(3) = 5
3. 1/12
Step-by-step explanation:
f(x) = 7e^x
lim x→0 7e^x =7
AS we can see from the graph it is around 7
We know that e^0 =1 so 7*1 =7
2. The limit does not exists, because approaching the limit from the left f(3-) is 3. Approaching the limit from the right f(3+) is 5. f(3) = 5 since there is a closed circle at 3 with a value of 5.
3. f(x) = x^ 1/2
We take the derivative
f'(x) = 1/2 x^ (-1/2)
= 1/2 1/( x^.5)
To find it at x=36
= 1/2 * 1/( 36^.5)
= 1/2 * 1/6
= 1/12
Since they are not equal, the limit does not exist.
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Answer:
x = -
Step-by-step explanation:
To find f(g(x)) substitute x = g(x) into f(x), that is
f(g(x))
= f(x + 1)
= 2(x + 1)² ← expand using FOIL
= 2(x² + 2x + 1) ← distribute
= 2x² + 4x + 2
To find g(f(x)) substitute x = f(x) into g(x), that is
g(f(x))
= g(2x²)
= 2x² + 1
----------------------------------------------------------
Equating gives
2x² + 4x + 2 = 2x² + 1 ( subtract 2x² + 1 from both sides )
4x + 1 = 0 ( subtract 1 from both sides )
4x = - 1 ( divide both sides by 4 )
x = -