f(x) = x ^ (3/2) Find an equation for the line tangent to the graph of f^ prime at the point (4, f^ prime (4))

Yaj143 Jan 30, 2021

#1**0 **

Close reading reveals that we are finding the equation of the tangent line of the derivative curve. That's a little unusual, but we can still do it.

Calculating the first- and second-order derivatives is fairly simple in this particular case.

\(f(x)=x^\frac{3}{2}\\ f'(x)=\frac{3}{2}x^\frac{1}{2}\\ f''(x)=\frac{3}{4}x^{-\frac{1}{2}}\)

f''(x) is a general equation for the slope of the tangent line of f'(x) given any x. Now, let's find it at the desired value, 4, in this case.

\(f''(4)=\frac{3}{4}*4^{-\frac{1}{2}}\\ f''(4)=\frac{3}{8}\)

This is the slope of the tangent line. In order to generate the equation of the line, however, we need one point that lies on the line. We will use the derivate curve to obtain the coordinates of one point.

\(x=4:\\ f'(4)=\frac{3}{2}*4^\frac{1}{2}\\ f'(4)=3\\ \text{Coordinate on Tangent Line: }(4,3)\)

I will use point-slope form of a line to construct the equation of the tangent line of the derivative curve.

\(y-y_1=f'(x_1)(x-x_1)\\ y-3=\frac{3}{8}(x-4)\\ y-3=\frac{3}{8}x-\frac{3}{2}\\ y=\frac{3}{8}x+\frac{3}{2}\)

This is the equation of the tangent line.

Guest Jan 30, 2021