Since the tangent line to function f(x) = 1 / sqrt(x) has to be parallel to the line x + 2y - 6 = 0, both needs to have the same slope.

Equation of the straight line can be rewritten in slope-intercept form to : y = -x/2 + 3. Hence the desired slope of our line is -1/2.

And derivative of f(x) will provide the slope of the line tangent to f(x)

d/dx(f(x)) = -1/2 * (x) ^ (-3/2)

setting it equal to -1/2

-1/2 * (x) ^ (-3/2) = -1/2 or x = 1 provides one of the conditions

Corresponding y coordinate can be found from the f(x) = 1/sqrt(x) equation, y = 1/sqrt(1) = 1

The slope of the desired line is equal to -1/2, and we now have a point on the desired line (1, 1), as well the point of tangency

Using the point-slope, equation to desired line can be obtained as follows. Given slope m = -1/2 and a point on the desired line,(x0, y0), the equation of the line is given by

(y - y0) = m(x - x0)

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

y = -1/2x + 3/2

2y + x - 3 = 0

The given line is :-

x + 2y - 6 = 0

or, y = -(1/2)x + 3 (the line is represented in slope intercept form)

Thus, the slope of the line = -(1/2)

Now, the tangent to the curve f(x) = 1/(x^(1/2)) is parallel to the above line

Thus, the slope of the tangent = slope of the line = -(1/2).......(1)

The given function is:-

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

differentiating both sides w.r.t 'x' we get

f'(x) = -(1/2)*[1/{x^(3/2)}]

Now, slope of the tangent = -(1/2)

Thus, -(1/2)*[1/{x^(3/2)}] = -(1/2)

or, x = 1

Putting the value of x =1 in the given equation of curve, we get

f(1) = y = 1

Hence the tangent passes through the point (1,1)

Thus, equation of the tangent at the point (1,1) and having slope = -(1/2) is :-

y - 1 = (-1/2)*(x - 1)

or, 2y - 2 = -x + 1

or, 2y + x - 3 = 0 is the equation of the tangent to the given curve at (1,1)

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now