(1 point) Find an equation for the line tangent to the graph of f(x) = sqrt(x) / √(3x−1) at the point (1, f(1) ). Y = ?

Answer 1

To write an equation for a line, you'll need a slope, and a point that the line goes through. Here you're given a point, (1, f(1)), and you know that the line needs to be tangent to this point. You may have learned this in class, but a derivative of a function will give you the rate of change, in this case the slope, of a line at a given point. So to get our slope, we'll need to find the derivative of your function F9x).

Because f(x) has a fraction, with x on the top and bottom, you'll need to use the quotient rule. The quotient rule is written like this:

f(x) = h(x)/g(x)

f'(x) = [g(x)h'(x) - h(x)g'(x)] / [g(x)]², where f(x) = original function

h(x) = numerator of fraction

g(x) = denominator of fraction

f'(x) = derivative of original function

g'(x) = derivative of numerator

h'(x) = derivative of denominator

When you first start using the quotient rule, I find it easier to write our your h(x), g(x) and their derivatives to start, then just plug everything into the equation above. I should note here that when taking a derivative, you can rewrite any roots as exponents to be able to use the power rule. So:

h(x) = √x

= x^1/2

h'(x) = (1/2)x^-1/2 using the power rule

g(x) = √(3x-1)

= (3x-1)^1/2

g'(x) = (1/2)(3x-1)^-1/2(3) using the chain rule

= (3/2)(3x-1)^-1/2

Now plug these into the quotient rule:

f'(x) = [g(x)h'(x) - h(x)g'(x)] / [g(x)]²

= [(3x-1)^1/2(1/2)x^-1/2 - x^1/2(3/2)(3x-1)^-1/2] / [(3x-1)^1/2]² Now there is a lot we can do to simplify this, but because we just want the slope, we can save some work by just plugging the x value of our point, (1,f(1)) and using a calculator to get the slope:

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

= -1 / (4√2) exact answer

= -0.177 to the thousandth decimal

Use this in the point slope form of an equation to get your equation. The point slope form looks like this:

y - y₁ = m(x - x₁) where y₁ = y coordinate of a point the line goes through

m = slope

x₁ = x coordinate of the same point the line goes through

Find f(1) to get your y₁:

f(1) = √1 / √(3(1)-1)

= √1 / √(3 - 1)

= √1 / √2

= 1/√2

Now plug everything into the point slope form:

y - 1/√2 = -1 / (4√2) (x - 1)

y = -x / (4√2) + 1 / (4√2) + 1/√2

= (-x + 1) / (4√2) + 1/√2)

Answer 2

y = sqrt(x)/ sqrt(3x-1)

= sqrt(x/(3x-1))

f(1) = sqrt( 1/2) = sqrt(2)/2

y' = (1/2)(x/(3x-1))^(-1/2) [ [3x-1 - 3x]/(3x-1)^2 ]

= (1/2) (x / (3x-1)^(-1/2) [ -1/(3x-1)^2 ]

f'(1) = (1/2) sqrt(2) (-1/4)

= -sqrt(2)/8

B = y - mx = sqrt(2)/2 - (-sqrt(2)/8)(1)

= sqrt(2)/2 + sqrt(2)/8

= (5/8) sqrt(2)

y = -sqrt(2)/8 x + (5/8) sqrt(2)

= (-sqrt(2))/8 [ x + 5]