What is the equation for the tangent line at x=3 when f(x)=1+x^(1/2) using the equation: f'(x)=lim(h to 0): [f(x+h)-f(x)]/h?? Thank you!!!

1 Answer
Feb 5, 2018

#x-2sqrt(3)y=-2sqrt(3)-3#

Explanation:

If #f(x)=1+x^(1/2)color(white)("xxx")=1+sqrt(x)#
then
#color(white)("XXXX")f'(x)=1/(2sqrt(x))#
[see below for derivation of this using the limit definition of derivative].

At #x=3#
#color(white)("XX")f(x=3)=1+sqrt(3)#
#color(white)("xXXXx")rarrcolor(white)("xx")"giving the point: "(x_0,y_0)=(3,1+sqrt(3))#
and
#color(white)("XX")f'(x=3)=1/(2sqrt(3))#
#color(white)("xXXXx")rarrcolor(white)("xx")"giving a slope of "m=1/(2sqrt(3))#

Using the slope-point form: #y-y_0=m(x-x_0)#
we get
#color(white)("XX")y-(1+sqrt(3))=(1/(2sqrt(3)))(x-3)#
which simplifies (in standard form) as:
#color(white)("XX")x-2sqrt(3)y=-2sqrt(3)-3#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using the limit definition to derive the slope (derivative)

#f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#

#color(white)("XX")=lim_(hrarr0)(sqrt(x+h)-sqrt(x))/h#

#color(white)("XX")=lim_(hrarr0)(sqrt(x+h)-sqrt(x))/h xx (sqrt(x+h)+sqrt(x))/(sqrt(x+h)+sqrt(x))#

#color(white)("XX")=lim_(hrarr0)((x+h)-(x))/(h(sqrt(x+h)+sqrt(x))#

#color(white)("XX")=lim_(hrarr0)1/(sqrt(x+h)+sqrt(x))#

#color(white)("XX")=1/(2sqrt(x))#