In the first Mean Value Theorem #f(b)=f(a)+(b-a)f'(c), a<c<b, f(x) =log_2 x, a=1 and f'(c)=1. How do you find b and c?

Answer 1

#b=1,2;c=1/ln2#

#f(b)=f(a)+(b-a)f'(c)#
#f(x)=log_2x#
Note that if #a=1#, then #f(a)=log_2 1=0#.
Also note that #f(x)=lnx/ln2# so #f'(x)=1/(xln2)#.
If #f'(c)=1#, then #1/(cln2)=1# so #c=1/ln2#.
Now we can also solve for #b#:
#f(b)=f(a)+(b-a)f'(c)#
We can replace #f(b)# with #log_2b#, #f(a)# with #0#, #a# with #1#, and #f'(c)# with #1#.
#log_2b=0+(b-1)(1)#
#log_2b=b-1#

This can be manipulated to show:

#b=2^(b-1)#

Or:

#2b=2^b#
This occurs at #b=1# and #b=2#.
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Answer 2

To find (b) and (c) in the first Mean Value Theorem equation (f(b) = f(a) + (b - a)f'(c)) where (a < c < b), and given that (f(x) = \log_2 x), (a = 1), and (f'(c) = 1), you can follow these steps:

  1. Substitute the given values into the equation: (f(a) = \log_2 1 = 0) since (\log_2 1 = 0). (f'(c) = 1).

  2. Solve the equation for (b): (f(b) = f(a) + (b - a)f'(c)) (f(b) = 0 + (b - 1)(1)) (f(b) = b - 1)

  3. Since (f(b) = \log_2 b), equate it to (b - 1): (\log_2 b = b - 1)

  4. Solve for (b) by finding the value that satisfies this equation.

  5. Once you find the value of (b), substitute it back into the equation to find (c).

  6. Given that (a = 1) and (b) is the value you found, choose (c) such that (1 < c < b).

This will provide the values of (b) and (c) that satisfy the conditions of the first Mean Value Theorem for the given function (f(x) = \log_2 x) and (f'(c) = 1).

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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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