The domain of f(x) is the domain of y=lnx, which is x>0.
Using product rule, f '(x) = 7 - 2lnx - 2 = 5 - 2lnx. Set f '(x) equal to 0 to find any critical values:
5 - 2lnx = 0
lnx = 2.5
x = e^2.5 ≈ 12.1825
Since f'(x) > 0 on the interval (0,e^2.5) and f '(x) < 0 on the interval (e^2.5, ∞), the function has a relative and absolute maximum at (e^2.5, f(e^2.5)) = (e^2.5, 2e^2.5).
f(x) = x - 3x⁻¹ + 6x⁻³
f ' (x) = 1 + 3x⁻² - 18x⁻⁴
Critical values are where f '(x) = 0 or is undefined. It's clearly undefined at x = 0, so we need to find where f '(x) = 0:
0 = 1 + 3x⁻² - 18x⁻⁴
Multiply by x⁴:
0 = x⁴ + 3x² - 18
0 = (x² + 6)(x²-3)
x = ±√3
Test the following intervals in f '(x) to determine where f(x) is increasing and decreasing. x = 0 is a vertical asymptote, but still needs to be included as a critical value since f(x) can change direction on either side of the asymptote:
(-∞, -√3):: test value x = -4 :: f '(x) > 0 → f(x) is increasing
(-√3, 0):: test value x = -1 :: f '(x) < 0 → f(x) is decreasing
(0,√3):: test value x = 1 :: f '(x) < 0 → f(x) is decreasing
(√3, ∞):: test value x = 4 :: f '(x) > 0 → f(x) is increasing
relative maximum at x = -√3, which is at the point (-√3, -2/√3)
relative minimum at x = √3, which is at the point (√3, 2/√3)
Comments
The domain of f(x) is the domain of y=lnx, which is x>0.
Using product rule, f '(x) = 7 - 2lnx - 2 = 5 - 2lnx. Set f '(x) equal to 0 to find any critical values:
5 - 2lnx = 0
lnx = 2.5
x = e^2.5 ≈ 12.1825
Since f'(x) > 0 on the interval (0,e^2.5) and f '(x) < 0 on the interval (e^2.5, ∞), the function has a relative and absolute maximum at (e^2.5, f(e^2.5)) = (e^2.5, 2e^2.5).
f(x) = x - 3x⁻¹ + 6x⁻³
f ' (x) = 1 + 3x⁻² - 18x⁻⁴
Critical values are where f '(x) = 0 or is undefined. It's clearly undefined at x = 0, so we need to find where f '(x) = 0:
0 = 1 + 3x⁻² - 18x⁻⁴
Multiply by x⁴:
0 = x⁴ + 3x² - 18
0 = (x² + 6)(x²-3)
x = ±√3
Test the following intervals in f '(x) to determine where f(x) is increasing and decreasing. x = 0 is a vertical asymptote, but still needs to be included as a critical value since f(x) can change direction on either side of the asymptote:
(-∞, -√3):: test value x = -4 :: f '(x) > 0 → f(x) is increasing
(-√3, 0):: test value x = -1 :: f '(x) < 0 → f(x) is decreasing
(0,√3):: test value x = 1 :: f '(x) < 0 → f(x) is decreasing
(√3, ∞):: test value x = 4 :: f '(x) > 0 → f(x) is increasing
relative maximum at x = -√3, which is at the point (-√3, -2/√3)
relative minimum at x = √3, which is at the point (√3, 2/√3)
These are relative (i.e., local) extrema. There is no absolute minimum (see http://www.wolframalpha.com/input/?i=y%3Dx-3%2Fx%2...
f(x) = 7x - 2x ln (x)
f'(x) = 7 - 2 ln(x) - 2x / x
f'(x) = 7 - 2 ln (x) -2 = 0
-2 ln (x) = -5
ln(x) = 5/2
x=e^(5/2) is a critical point
f''(x) = -2/x < 0 when x=e^5, so f(x) has a local global maximum when x=e^(5/2)
The global maximum is f(e^(5/2)) = 7 [e^(5/2)] - 2 [e^(5/2)] ln [e^(5/2) ] = 24.365
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f(x) = x - 3/x + 6/x^3
f(x) = x - 3 x^(-1) + 6 x^(-3)
f'(x) = 1 - 3(-1) x^(-2) + 6(-3) x^(-4)
f'(x) = 1 + 3/x^2 -18/x^4 = 0
multiply everything by x^4
x^4+3x^2-18 = 0
Let y=x^2
y^2-3y-18=0
(y+6)(y-3) = 0
y = -6 ; y= 3
x^2=-6 (not possible)
x^2 = 3
x = 屉 3
f'(x) = 1 + 3/x^2 -18/x^4
f'(x) = 1 + 3 x^(-2) - 18 x^(-4)
f''(x) = 3 (-2) x^(-3) - 18 (-4) x^(-5)
f''(x) = -6 /x^3 +72 /x^5
When x= â 3 , f''(x) = - 6 / (â 3)^3 + 72 / (â 3)^5 = 3.464 > 0
f has a 'local minimum'
The local minimum is f(â 3)
substitute x=â 3 and evaluate x-3/x+6/x^3 = 3.464
The local minimum is 3.464
There is no absolute minimum