F(9) if f(x) = 5x - 16
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Critical points is where the derivative (slope) is zero or does not exist. So to do this we have to find the derivative of our function:
So we apply chain rule:
=
Set our first derivative to zero and solve for x:
3(x^2 - 1) * 2x = 0
So we can see that (by plugging in) 0, -1 and 1 makes our solution true
So our critical value is x = 0, x = -1, x = 1
So we apply chain rule:
=
Set our first derivative to zero and solve for x:
3(x^2 - 1) * 2x = 0
So we can see that (by plugging in) 0, -1 and 1 makes our solution true
So our critical value is x = 0, x = -1, x = 1
Let x represent the larger number.
1/(x -28) +4/x = 1/6
Multiplying by the product of denominators, we have
6x +24(x -28) = x(x -28)
In standard form, this is
x^2 -58x +672 = 0
(x -42)(x -16) = 0
The two numbers are (-12 and 16) or (14 and 42).
1/(x -28) +4/x = 1/6
Multiplying by the product of denominators, we have
6x +24(x -28) = x(x -28)
In standard form, this is
x^2 -58x +672 = 0
(x -42)(x -16) = 0
The two numbers are (-12 and 16) or (14 and 42).