(x^2-x+4)/(x-1)
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differentiate using quotient rule, [(x-1)(2x-1) - (x^2-x +4)(1)]/ (x-1)^2
ie [2x^2 -3x +1 -x^2 +x -4]/(x-1)^2 --->0 ie [x^2 -2x -3]/(x-1)^2 --->0 ie (x-3)(x+1)/(x-1)^2 --->0 x -->3 or x--->-1 consider the sign of (x-3)(x+1)/(x-1) ^2 on either side of these values of x.
thus when x --> 2.9 (-)(+)/(+) ---> -
when x--->3.1 (+)(+)/(+) ---> + that is changes from negative to positive as x increases so x-->3 is a minimum
when x --->-1.1 (-)(-)/(+) --->+
when x --->-0.9 (-)(+)/(+) --->- ie changes from positive to negative as x increases
so x--->-1 is a maximum
ie [2x^2 -3x +1 -x^2 +x -4]/(x-1)^2 --->0 ie [x^2 -2x -3]/(x-1)^2 --->0 ie (x-3)(x+1)/(x-1)^2 --->0 x -->3 or x--->-1 consider the sign of (x-3)(x+1)/(x-1) ^2 on either side of these values of x.
thus when x --> 2.9 (-)(+)/(+) ---> -
when x--->3.1 (+)(+)/(+) ---> + that is changes from negative to positive as x increases so x-->3 is a minimum
when x --->-1.1 (-)(-)/(+) --->+
when x --->-0.9 (-)(+)/(+) --->- ie changes from positive to negative as x increases
so x--->-1 is a maximum
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Let f(x) = (x^2-x+4)/(x-1)
Take the derivative using the Quotient Rule to get f ' (x) = (x^2 - 2x - 3)/[(x - 1)^2]
The local extremes occur when f ' (x) = 0, so set f ' (x) = 0.
Hence, (x^2 - 2x - 3)/[(x - 1)^2] = 0, but a fraction = 0 only if the numerator is 0.
Therefore, solve x^2 - 2x - 3 = 0 and factor to get (x - 3)(x + 1) = 0.
Thus, the local extremes occur at x = 3 and x = -1.
Draw a number line/chart for the zeros of f ' . RECALL that f is increasing when f ' >0 and decreasing when f ' < 0.
On the left of x = -1, we get that f ' > 0 since x -3 and x + 1 with both be negative. If x is in between -1 and 3, we get that x -3 < 0 and x + 1 >0 therefore, f ' is negative there. Thus, there's a local max at x = -1.
If x >3, then x -3, and x + 1 are both positive so f ' is positive there. Therefore, a local min at x =3.
Take the derivative using the Quotient Rule to get f ' (x) = (x^2 - 2x - 3)/[(x - 1)^2]
The local extremes occur when f ' (x) = 0, so set f ' (x) = 0.
Hence, (x^2 - 2x - 3)/[(x - 1)^2] = 0, but a fraction = 0 only if the numerator is 0.
Therefore, solve x^2 - 2x - 3 = 0 and factor to get (x - 3)(x + 1) = 0.
Thus, the local extremes occur at x = 3 and x = -1.
Draw a number line/chart for the zeros of f ' . RECALL that f is increasing when f ' >0 and decreasing when f ' < 0.
On the left of x = -1, we get that f ' > 0 since x -3 and x + 1 with both be negative. If x is in between -1 and 3, we get that x -3 < 0 and x + 1 >0 therefore, f ' is negative there. Thus, there's a local max at x = -1.
If x >3, then x -3, and x + 1 are both positive so f ' is positive there. Therefore, a local min at x =3.
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f(x) = (x^2-x+4)/(x-1)
f'(x) = [(x-1)(2x-1)-(x^2-x-4)]/(x-1)^2 = 0 for extrema
2x^2-3x+1-x^2+x-4 = 0
x^2-2x-3 = 0
(x-3)(x+1) = 0
x = 3 and -1
f(3) = 5
f(-1) = -3
When x = 3, f''(x) > 0 so is a local minimum
When x = -1, f''(x) < 0 so is a local maximum
Local maximum is (-1,-3)
Local minimum is (3,5)
f'(x) = [(x-1)(2x-1)-(x^2-x-4)]/(x-1)^2 = 0 for extrema
2x^2-3x+1-x^2+x-4 = 0
x^2-2x-3 = 0
(x-3)(x+1) = 0
x = 3 and -1
f(3) = 5
f(-1) = -3
When x = 3, f''(x) > 0 so is a local minimum
When x = -1, f''(x) < 0 so is a local maximum
Local maximum is (-1,-3)
Local minimum is (3,5)