How do you find the equation of the line tangent to the graph of #(2x+3)^(1/2)# at the point x=3?

Question

Samantha Adkison · Accepted Answer

#y=x/3+2#

First use the chain rule to get the derivative:

Let:

#y=(2x+3)^(1/2)#

#y'=1/2(2x+3)^(-1/2)xx2#

=#(1)/(sqrt(2x+3))#

If #x=3#:

#y'=(1)/(sqrt(9))=1/3#

The equation of the line is of the form:

#y=mx+c#

So #m=1/3#

If #x=3# then #y=sqrt((2xx3)+3)=3#

So #3=1/3xx3+c#

#c=3-1=2#

So the equation#rArr#

#y=x/3+2#

Emma Aldridge · Answer

undefined To find the equation of the line tangent to the graph of (2x+3)^(1/2) at the point x=3, we need to find the derivative of the function and evaluate it at x=3.

First, we find the derivative of (2x+3)^(1/2) using the power rule of differentiation. The derivative is given by:

dy/dx = (1/2)(2x+3)^(-1/2)(2)

Simplifying this expression, we get:

dy/dx = (2)/(2x+3)^(1/2)

Next, we evaluate the derivative at x=3:

dy/dx = (2)/(2(3)+3)^(1/2)
       = (2)/(6+3)^(1/2)
       = (2)/(9)^(1/2)
       = (2)/3

So, the slope of the tangent line at x=3 is 2/3.

Now, we have the slope of the tangent line and the point (3, (2(3)+3)^(1/2)) on the graph. We can use the point-slope form of a line to find the equation of the tangent line:

y - y1 = m(x - x1)

Plugging in the values, we get:

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

Simplifying further:

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

This is the equation of the line tangent to the graph of (2x+3)^(1/2) at the point x=3.