f(x) = x - 1/(1+5x)
= x - (1+5x)^-1
f'(x) = 1 + (1+5x)^-2 (5)
= 1 + 5/(1+5x)^2
the slope of x-2y=2 is 1/2
so
1 + 5/(1+5x)^2 = 1/2
5/(1+5x)^2 = -1/2
(1+5x)^2 = -10
A square of something cannot be negative, so
either there is no solution or you made an error in typing.
Was the equation perhaps
f(x) = (x-1)/(1+5x) ?
If f(x)=[x−1/(1+5 x)], there are two points on the graph of
y=f(x) at which the tangent lines are parallel to the line x−2y=2.
Type in the x values of
these points. Type the smaller number first, then the larger number.
2 answers
yeah that's what I meant
1 answer