# How do you find the point c in the interval #1<=x<=4# such that f(c) is equation to the average value of #f(x)=abs(x-3)#?

The leftmost triangle has base 3-1 = 2 and height of |1 - 3| = 2 so it has area 2.

The rightmost triangle has base 4 - 3 = 1 and height of |4 - 3| = 1, so it has area 1/2.

The function takes that value at two points:

As is hopefully clear in the above example, one uses calculus (here geometry) in order to find the average value and just algebra to find the values. We know that any continuous function has to cross its average value at least once within any interval, but there's no easy/formulaic way to find where that is true without doing algebra.

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To find the point (c) in the interval ([1, 4]) such that (f(c)) is equal to the average value of (f(x) = |x - 3|), we first need to find the average value of (f(x)) over the interval ([1, 4]).

The average value of a function (f(x)) over an interval ([a, b]) is given by:

[ \text{Average value} = \frac{1}{b - a} \int_a^b f(x) , dx ]

For (f(x) = |x - 3|) over the interval ([1, 4]), we have:

[ \text{Average value} = \frac{1}{4 - 1} \int_1^4 |x - 3| , dx ]

We can split the integral into two parts based on the definition of the absolute value function:

[ \int_1^4 |x - 3| , dx = \int_1^3 (3 - x) , dx + \int_3^4 (x - 3) , dx ]

Now, integrate each part separately:

[ = \left[ \frac{-(3 - x)^2}{2} \right]_1^3 + \left[ \frac{(x - 3)^2}{2} \right]_3^4 ]

[ = \left[ \frac{-(3 - 3)^2}{2} - \frac{-(3 - 1)^2}{2} \right] + \left[ \frac{(4 - 3)^2}{2} - \frac{(3 - 3)^2}{2} \right] ]

[ = \left[ 0 - \frac{-2^2}{2} \right] + \left[ \frac{1^2}{2} - 0 \right] ]

[ = \left[ 0 - 2 \right] + \left[ \frac{1}{2} \right] ]

[ = -2 + \frac{1}{2} ]

[ = -\frac{3}{2} ]

So, the average value of (f(x)) over the interval ([1, 4]) is (-\frac{3}{2}).

To find the point (c) such that (f(c)) equals (-\frac{3}{2}), we need to solve the equation (|c - 3| = -\frac{3}{2}).

However, since the absolute value of any real number is non-negative, there is no real number (c) such that (|c - 3| = -\frac{3}{2}). Thus, there is no point (c) in the interval ([1, 4]) such that (f(c)) is equal to the average value of (f(x) = |x - 3|).

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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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