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.
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) ?
yeah that's what I meant
To find the x-values of the points on the graph of y=f(x) where the tangent lines are parallel to the line x - 2y = 2, we can follow these steps:
Step 1: Find the derivative of f(x):
The derivative of f(x) can be found using the quotient rule. Let's differentiate f(x) to get f'(x):
f(x) = [x - 1/(1 + 5x)]
Using the quotient rule, we have:
f'(x) = [1(1 + 5x) - (x - 1)(5)] / (1 + 5x)^2
= [1 + 5x - 5x + 5] / (1 + 5x)^2
= [6] / (1 + 5x)^2
= 6 / (1 + 5x)^2
Step 2: Find the slope of the line x - 2y = 2:
The given line is x - 2y = 2. To find the slope, we need to rewrite it in slope-intercept form, y = mx + b:
x - 2y = 2
-2y = -x + 2
y = (1/2)x - 1
From this equation, we can see that the slope is 1/2.
Step 3: Set the derivative equal to the slope:
Set f'(x) = 1/2 and solve for x:
6 / (1 + 5x)^2 = 1/2
1 + 5x = √12 (Taking the square root on both sides)
5x = √12 - 1
x = (√12 - 1)/5
Step 4: Find the x-values of the points on the graph:
To find the x-values of the points on the graph where the tangent lines are parallel to the line x - 2y = 2, we substitute the slope into f'(x) and solve for x:
(√12 - 1)/5 = (√12 - 1)/(5(1 + 5x)^2)
Cross multiply for simplicity:
(√12 - 1)(5(1 + 5x)^2) = 5
By solving this equation, we can find the x-values of the points on the graph. However, the equation involves a quadratic term, so the algebraic manipulation can be complex.