Equation for the line tangent to the graph?

Question

Find an equation for the line tangent to the graph of

f(x)= sqrt(x)/ 2x−3 at the point (2,f(2)) .

y = ?

Answer 1

The function f(x) = √(x)/(2x - 3) and the point is (2, f(2))).

f(2) = √(2)/(2*2 - 3) = √(2)/(4 - 3) = √2.

So, the point is (2, √2).

f(x) = √(x)/(2x - 3) = √x * (2x - 3)^{- 1}.

Apply product rule : (uv) ' = uv ' +  vu '.

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

Apply power rule : (xⁿ) ' = nx^{n - 1}.

Applu formula : [√x] ' = 1/(2√x).

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

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

when, x = 2,

f '(2) = [ - √2/(2*2 - 3)²] + [1/(2√2(2*2 - 3))]

= [ - √2/(4 - 3)²] + [1/(2√2(4 - 3))]

= [ - √2/(1)²] + [1/(2√2(1))]

= [ - √2] + [1/(2√2)]

= - 3/(2√2).

⇒ f ' (x) = - 3/(2√2), this is the slope of the tangent line.

Slope - intercept form line equation is y = mx + b, where m is slope and b is y-intercept.

m = - 3/(2√2).

Now, the line equation is y = (- 3/(2√2))x + b.

Find the y - intercept by substituting the given point in the line equation say (x, y) = (2, √2).

√2 = [- 3/(2√2)](2) + b

b = √2 + 3/√2

b = (2 + 3)/√2

⇒ b = 5/√2.

Therefore, the tangent line equation is y = [- 3/(2√2)]x + (5/√2).