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# 4[x]=x+{x}[]= greatest int.; {}= fractional partsolve for exact values of x.

10 years ago

The slope of the secant line that passes through the endpoints is:
m = ?y/?x
= [f(6) - f(1)]/(6 - 1)
= [(6 + 4/6) - (1 + 4/1)]/5
= 1/3.

By the MVT, there does exist a c ∈ (1, 6) such that f'(c) = 0.

By re-writing f(x) = x + 4/x as x + 4x^(-1) and differentiating gives:
f'(x) = 1 + (4)(-1)x^(-2)
= 1 - 4/x^2
= (x^2 - 4)/x^2.

Then, f'(c) = (c^2 - 4)/c^2.

Setting f'(c) = 1/3:
(c^2 - 4)/c^2 = 1/3
==> 3(c^2 - 4) = c^2, by cross multiplying
==> 2c^2 - 12 = 0
==> c = -√6 and c = √6.

Then, since 1 < c < 6, c = √6 ≈ 2.449.

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