Help with critical points in Calc
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Help with critical points in Calc

[From: ] [author: ] [Date: 12-02-29] [Hit: ]
(x+5)^2 ≥ 0 but you cant let x = -5 which makes the denominator becomes 0,There is no critical point!NONE.-take the derivative and set it equal to zero to find the critical points. in this case df(x)/dx = 23/(x+5)^2 which never will equal zero, so there are no critical points.......
Suppose that

f(x) = (4x − 3)/(x + 5)
.


Find all critical values of f . If there are no critical values, enter None. If there are more than one, enter them separated by commas.
Critical value(s) =

-
f('x) = 4/(x + 5) - (4x − 3)/(x + 5)^2

where 4/(x + 5) - (4x − 3)/(x + 5)^2 = 0

4(x+5)/(x + 5)^2 - (4x − 3)/(x + 5)^2 = 0

[4(x+5) - (4x − 3)] /(x + 5)^2 = 0

[4x + 20 - 4x + 3)] /(x + 5)^2 = 0

23/(x + 5)^2 = 0

(x+5)^2 ≥ 0 but you can't let x = -5 which makes the denominator becomes 0,

There is no critical point! NONE.

-
take the derivative and set it equal to zero to find the critical points. in this case df(x)/dx = 23/(x+5)^2 which never will equal zero, so there are no critical points.

-
f '(x) > 0 for any values of x in its domain ---> no critical value
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