#x^3+24x-16# [0,4] verify mean value theorem?

Answer 1

We seek to verify the Mean Value Theorem for the function

# f(x) = x^3+24x-16# on the interval #[0,4]#
The Mean Value Theorem, tells us that if #f(x)# is differentiable on a interval #[a,b]# then #EE \ c in [a,b]# st:
# f'(c) = (f(b)-f(a))/(b-a) #
So, Differentiating wrt #x# we have:
# f'(x) = 3x^2 + 24 #
And we seek a value #c in [0,4]# st: # f'(c) = (f(4)-f(0))/(4-0) #
# :. 3c^2 + 24 = ((64+96-16)-(0+0-16))/(4-0) #
# :. 3c^2 + 24 = 160/4 #
# :. 3c^2 + 24 = 40 #
# :. 3c^2 = 16 #
# :. c^2 = 16/3 #
# :. c = +- (4sqrt(3))/3 #
And we require that #c in [0,4]#, so we choose #c=(4sqrt(3))/3#
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Answer 2

To verify the Mean Value Theorem for the function ( f(x) = x^3 + 24x - 16 ) over the interval ([0,4]), follow these steps:

  1. Verify that the function ( f(x) ) is continuous on the closed interval ([0,4]). This function is a polynomial function, and polynomial functions are continuous for all real numbers.

  2. Verify that the function ( f(x) ) is differentiable on the open interval ((0,4)). Since ( f(x) ) is a polynomial function, it is differentiable for all real numbers.

  3. Calculate the average rate of change of ( f(x) ) over the interval ([0,4]) using the formula: [ \frac{f(b) - f(a)}{b - a} ] where ( a = 0 ) and ( b = 4 ).

  4. Calculate the derivative of ( f(x) ) with respect to ( x ), ( f'(x) ), which is ( 3x^2 + 24 ).

  5. Find the value of ( c ) in the open interval ((0,4)) such that ( f'(c) ) is equal to the average rate of change of ( f(x) ) over ([0,4]), by solving the equation: [ f'(c) = \frac{f(4) - f(0)}{4 - 0} ]

  6. Once you find the value of ( c ), verify that it lies within the interval ((0,4)).

  7. If ( c ) lies within the interval ((0,4)), then the Mean Value Theorem is verified for the function ( f(x) ) over the interval ([0,4]), and the theorem states that there exists at least one ( c ) such that ( f'(c) = \frac{f(4) - f(0)}{4 - 0} ).

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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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