The figure below shows the graph f(x) = sqrt(x) and it’s tangent line at x=a. Find the value of a.

Question

The tangent line goes through the point (0,1) and is tangent at the point (a, sqrt(a)). I’m having trouble finding a numerical value for a.

Answer 1

A line tangent to a function at a point has a slope equal to the derivative of the function.

$f^{\prime }(x)=\frac{1}{2\sqrt{x}}$

$f^{\prime }(a)=\frac{1}{2\sqrt{a}}$

We know the slope is 1/(2*sqrt(a)). We know (0,1) and (a,sqrt(a)) are points on the tangent line.

Say the line has an equation of the form (y-y_1)=m*(x-x_1)

m=delta_y/delta_x=(y_2-y_1)/(x_2-x-1)=(sqrt(a)-1)/(a-0)=(sqrt(a)-1)/a

(sqrt(a)-1)/a=1/(2*sqrt(a))

Solving for a:

2a-2sqrt(a)=a

a=2sqrt(a)

sqrt(a)=2

a=4 only.