How do you find the point c in the interval #-1<=x<=1# such that f(c) is equation to the average value of #f(x)=x^2-x#?

Answer 1

Find the average value. Set the function equal to that value. Solve the equation in the interval. #c = 1/2 - sqrt21/6#

Average value on #[-1,1]#:
#bar(y) = 1/(1-(-1)) int_-1^1 (x^2-x) dx#
# = 1/2 [x^3/3-x^2/2]_-1^1#
# = 1/3#

Solve

#x^2-x = 1/3#
#3x^2-3x-1=0#
#x = (3+-sqrt(9+12))/6#
# = 1/2 +- sqrt21/6#
#5 < sqrt21 < #6, so #1/2 < sqrt21/6 < 1#. Thus
#1/2 + sqrt21/6# is outside the interval and
#1/2 - sqrt21/6# is in the interval #[-1,1]#.
#c = 1/2 - sqrt21/6#
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Answer 2

To find the point c c in the interval 1x1-1 \leq x \leq 1 such that f(c) f(c) is equal to the average value of f(x)=x2x f(x) = x^2 - x , you would first calculate the average value of f(x) f(x) over the interval [1,1][-1, 1] using the formula:

Average value=1baabf(x)dx\text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx

where a a and b b are the lower and upper bounds of the interval, respectively. Then, you would set f(c) f(c) equal to this average value and solve for c c .

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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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