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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.
To check if a piecewise defined function is continuous, you need to check how the pieces "glue" together when you step from one domain to the other.
So, the question is: what happens at x=3? If you reach x=3 from values slightly smaller than 3, you obey the rule f(x)=log(3x). So, as you approach 3, you get values closer and closer to
Similarly, if you reach x=3 from values slightly greater than 3, you obey the rule f(x)=(4-x)log(9). So, as you approach 3, you get values closer and closer to
So, the function is continuous at x=3, because both pieces approach log(9) as x approaches 3.