1. Solve for x. Show your work. 2x-1/2=3-x
2 answers:
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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
- The function should be continuous in [0,2]
- The function should be differentiable in (1,2)
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.
The integrals over B and T will be positive. Keeping fixed,
is strictly increasing over D as
increases, so the integrals over
(i.e. the bottom/top left quadrants of D) is negative but the integrals over
are *more* positive.
The integrals over R and L are zero. If we take , then
, which is to say
is symmetric across the
-axis. For the same reason, the integral over all of D is also zero.