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Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = 3x2 − 2x + 1, [0, 2] Yes, it does not matter if f is continuous or differentiable, every function satisfies the Mean Value Theorem. Yes, f is continuous on [0, 2] and differentiable on (0, 2) since polynomials are continuous and differentiable on . No, f is not continuous on [0, 2]. No, f is continuous on [0, 2] but not differentiab

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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) =[tex]3x^2 -2x +1[/tex]

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.

   

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