Answer:
g(x) = 1/2*(4)^(–x) and
g(x) =1/2*(1/4)^(x)
Please, see attached picture.
Step-by-step explanation:
Your full question is attached in the picture below
To easily solve this problem, we can graph each equation and see, which one represents a reflection of the function over the y axis.
See, second image.
The answers are
g(x) = 1/2*(4)^(–x) and
g(x) =1/2*(1/4)^(x)
Answer:
Yes , function is continuous in [0,2] and is differentiable (0,2) since polynomial function are continuous and differentiable
Step-by-step explanation:
We are given the Function
f(x) =
The two basic hypothesis of the mean valued theorem are
upon checking the condition stated above on the given function
f(x) is continuous in the interval [0,2] as the functions is quadratic and we can conclude that from its graph
also the f(x) is differentiable in (0,2)
f'(x) = 6x - 2
Now the function satisfies both the conditions
so applying MVT
6x-2 = f(2) - f(0) / 2-0
6x-2 = 9 - 1 /2
6x-2 = 4
6x=6
x=1
so this is the tangent line for this given function.
