Question

At what point on the graph of f(x)=2x-(1/x) is the tangent line parallel to the line 2y-12=7?

Answer 1

you dropped an x

Answer 2

oh here it is:
\[2y-12x=7\] i found it

Answer 3

slope of the line is what, 6?

Answer 4

it was hiding under the parsley

Answer 5

f'(x) = 6 then

Answer 6

sage, rosemary and thyme

Answer 7

so from there, you take the derivative?

Answer 8

yes, take the derivative of your f(x), since that is the equation that will give you the slope of the tangent line to any point on it

Answer 9

and equate it to 6

Answer 10

f(x)=2x-(1/x)

f'(x) = 2 + 1/x^2 right?

Answer 11

6 = 2 + 1/x^2

4 = 1/x^2

x^2 = 1/4 ... yada yada

Answer 12

1 + 1 = 2 and 3 x 3 = 9, I hope this helps you.

Answer 13

At what point on the graph of f(x)=2x-(1/x) is the tangent line parallel to the line 2y-12x=7?

The line \(2y-12x=7\) can be re-written as \(y = 6x - 7/2\). The slope of the line is therefore 6.

We want to find the values of x for which \(f'(x) = 6\).

Since \(f(x)=2x-(1/x)\) that means \(f'(x) = 2 +1/x^2\)

Solving \(6 = 2 + 1/x^2\) gives us \(\pm 1/2\). Those are the values of for which \(f(x)\) have slope 6.

Substitute those values back into the original function \(f(\pm 1/2)\) to obtain \((1/2, -1), (-1/2, 1)\). So now you write the equation of TWO lines -- each with slope 6, but one line goes through the (1/2, -1) and the other line goes through (-1/2, 1).

Glorious picture attached, made in Geogebra, of course. :D