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Given one to one f(6)=3, f'(6)=5, find tangent line to y=f⁻¹(x) at (3,6)

OK, we know

f⁻¹(3)=6

The inverse function is the reflection in y=x.  So slopes, i.e. the derivative will be the reciprocal.  We know the derivative of f at 6 is 5, so the derivative of f⁻¹  at y=6 is 1/5, which corresponds to x=3.

f⁻¹ ' (3) = 1/5

That slope through (3,6) is the tangent line we seek:

y - 6 = (1/5) (x-3)

That's the tangent line.

y = x/5 + 27/5

First, find the inverse function.

Note that an inverse is a reflection; y = x

The derivative given will be the reciprocal and when f is at 6, it equals 5. So, when it is at y = 6 it is at x = 1/5 and corresponds to 3.

f^-1(3) = 1/5

The point (3, 6) is the point in which the slope goes through. We can write all this information in slope-intercept form.

y = 1/5x + 27/5

Best of Luck1

other answer is correct. i had used the radius of the radius somehow. a website: