Suppose f(x)=√(x+4)

Question

a) Use the definition of derivative to find the slope of the tangent line to the cure at x = 2.

b) Find the equation of the tangent line to the curve at x = 2.

Answer 1

f(x+h) = sqrt( x+h+4)

f(x+h) - f(x) = sqrt(x+h+4) - sqrt(x+4)

difference quotient is over h. Rationalizes the numerator

by multiplying the top and bottom by sqrt(x+h+4) + sqrt(x+4)

(x+h+4) - (x+4)]/ { h[ sqrt(x+h+4) + sqrt(x+4)}

x+4 cancels, then the h cancels

the difference quotient becomes

1/ { sqrt(x+h+4) + sqrt(x+4)}

as h tends to zero, the difference quotient becomes

(1/2) (x+4)^(-1/2) which agrees with the power rule

For x = 2, the slope of the tangent line is (1/2) (1/ sqrt(6) = (1/2) * sqrt(6)/6 = sqrt(6)/12

f(2) = sqrt(6)

so the tangent line has slope sqrt(6)/12 and eats (2, sqrt(6))

B = y - mx = sqrt(6) - (sqrt(6)/12 * 2

= sqrt(6) = sqrt(6)/6

= (5/6)*sqrt(6)

y = sqrt(6)/12 X + (5/6) sqrt(6) = sqrt(6)X/12 + (10/12)*sqrt(6)

= sqrt(6)/12 ( X+10)