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.
What is the intersection of the three sets: A = {0, 2, 3, 6, 8}, B = {2, 3, 6, 8, 9}, and C = {1, 2, 4, 8, 9}? A. {2, 8, 9} B. {
Ugo [173]
Answer:
{2,8}
Step-by-step explanation:
This is the same thing as asking what element (in this case what number) is in all 3 sets.
0 isn't in all 3 sets because it isn't in B.
2 is in all 3 sets
3 isn't because it isn't in C
4 isn't in A.
6 isn't in C.
8 is in all 3 sets.
9 isn't in A
So the elements that are in the 3 sets are {2,8}.
Remember, you can do ANYTHING to an equaiton as long as you do it to BOTH SIDES
we can try to get k by itself bymaking that -17 into 0 since k+0=k
-17+17=0 right so
K-17=-12
add 17 to BOTH SIDES
K+17-17=17-12
K+0=5
K=5
Answer:D) SAS
Step-by-step explanation:
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