Asked by Sandy
Determine the critical numbers for
y=sq root of (x^2 - 2x -15)
y=sq root of (x^2 - 2x -15)
Answers
Answered by
Reiny
or
y = √((x-5)(x+3))
This function is only defined for x≥5 or x≤-3
(for other values the inside of the square root would be negative)
dy/dx = (1/2)(x^2-2x-15)^(-1/2) (2x-2)
= (x-1)/√(x^2-2x-15)
setting this equal to zero to get a max/min results
in x = 1
BUT, that lies in the restricted domain, so there is no max/min.
There are two x-intercepts, at
(-3,0) and (5,0)
There are no y-intercepts, (let x=0, to get y = √-15)
If you want to find the second derivative, you will find there are no points of inflection.
the graph basically consists of 2 "wings", one rising to the right in the first quadrant from (5,0), the other rising to the left in the 2nd quad from (-3,0)
y = √((x-5)(x+3))
This function is only defined for x≥5 or x≤-3
(for other values the inside of the square root would be negative)
dy/dx = (1/2)(x^2-2x-15)^(-1/2) (2x-2)
= (x-1)/√(x^2-2x-15)
setting this equal to zero to get a max/min results
in x = 1
BUT, that lies in the restricted domain, so there is no max/min.
There are two x-intercepts, at
(-3,0) and (5,0)
There are no y-intercepts, (let x=0, to get y = √-15)
If you want to find the second derivative, you will find there are no points of inflection.
the graph basically consists of 2 "wings", one rising to the right in the first quadrant from (5,0), the other rising to the left in the 2nd quad from (-3,0)
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