So, f[x] = 1/4x^2 - 1/2Ln(x)
<span>thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x </span>
<span>thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) </span>
<span>thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 </span>
<span>thus Sqrt[...] = (x/2 + 1/2x) </span>
The center of the circle is (5,3)
Note that √2 + √2 = 2√2
2/(2√2) = (√2)/2 = 0.707 (simplified)
0.707 is your answer
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F(x) = 2x - 4
f(2 ≤ x) = 2(2 ≤ x) - 4
f(x ≥ 2) = 2(x ≥ 2) - 4
f(x ≥ 2) = 2(x) ≥ 2(2) - 4
f(x ≥ 2) = 2x ≥ 4 - 4
f(x ≥ 2) = 2x ≥ 0
f(x ≥ 2) = x ≥ 0
f(x) = 2x - 4
f(x ≤ 6) = 2(x ≤ 6) - 4
f(x ≤ 6) = 2(x) ≤ 2(6) - 4
f(x ≤ 6) = 2x ≤ 12 - 4
f(x ≤ 6) = 2x ≤ 8
f(x ≤ 6) = x ≤ 4
Hello,
-5x-15>10+20x
==>-15-10>20x+5x
==>-25>25x
==>-1>x
==>x<-1
