Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval if #f(x) = 2x^2 − 3x + 1# and [0, 2]?

Answer 1
The Mean Value Theorem says that if a function #f(x#) is differentiable over a range #[x_1,x_2]# then there exists a value #c# within the range #[x_1,x_2]# such that
#f'(c) = (f(x_2) - (f(x_1)))/(x_2-x_1)# (The right hand side of this is often referred to as the secant).
For the given function #f(x) = 2x^2-3x+1# using #x_1=0# and #x_2=2#
#(f(2) - f(0))/(2-0)#
#=3/2#
#f'(x) = 4x-3#
If we (temporarily) assume the Mean Value Theorem holds then there must be a value #c# such that #4c - 3 = 3/2# which solves as #c=9/8#

Testing this back in the equation of the Mean Value Theorem demonstrates that the theorem holds for the given equation.

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Answer 2

Yes, the function ( f(x) = 2x^2 - 3x + 1 ) satisfies the hypotheses of the Mean Value Theorem on the interval ([0, 2]).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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