I don't see a table but I can give you the means to answer it yourself. The inverse function is represented by this:
where k is your constant. You are given a k value of 4. If you solve this for k then you will get xy=4. In your tables, multiply your x value by your y value within your coordinate points and if you get a product of 4 each time you multiply x by y, then that table is your answer.
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.
Answer:
2/5
Step-by-step explanation:
5x+6y=11
as we have to find x coordinate
while y coordinate 3/2 is already given in question
so only put value of y coordinate in given equation to get x coordinate value
5x+6y=11
5x+6*(3/2)=11
5x+3*3=11
5x+9=11
5x=11-9
5x=2
x=2/5
The only correct statement is that ratios are not fractions.
