# What is the value of c for which the instantaneous rate of change of f at #x=c# is the same as the average rate of change f over [1,4] if #f(x)=x+lnx#?

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To find the value of ( c ) for which the instantaneous rate of change of ( f ) at ( x=c ) is the same as the average rate of change of ( f ) over ([1,4]), we need to equate the derivative of ( f(x) ) at ( x=c ) with the average rate of change of ( f(x) ) over ([1,4]).

The derivative of ( f(x) = x + \ln(x) ) is ( f'(x) = 1 + \frac{1}{x} ).

The average rate of change of ( f ) over ([1,4]) is ( \frac{f(4) - f(1)}{4 - 1} ).

So, we have:

[ f'(c) = \frac{f(4) - f(1)}{4 - 1} ]

Substitute the expressions for ( f'(x) ), ( f(4) ), and ( f(1) ):

[ 1 + \frac{1}{c} = \frac{4 + \ln(4) - (1 + \ln(1))}{3} ]

Simplify:

[ 1 + \frac{1}{c} = \frac{4 + \ln(4)}{3} ]

[ \frac{1}{c} = \frac{4 + \ln(4)}{3} - 1 ]

[ \frac{1}{c} = \frac{4 + \ln(4) - 3}{3} ]

[ \frac{1}{c} = \frac{\ln(4) + 1}{3} ]

[ c = \frac{3}{\ln(4) + 1} ]

Therefore, the value of ( c ) for which the instantaneous rate of change of ( f ) at ( x=c ) is the same as the average rate of change of ( f ) over ([1,4]) is ( c = \frac{3}{\ln(4) + 1} ).

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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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