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
Answer: About 3.06
Step-by-step explanation:
We can use trigonometry functions to solve for AC. Let the ?, representing AC, be "x" in our mathematical work.
Since we have the hypotenuse and x is adjacent to the angle given, I am going to use cosine.
cos(θ) = 
cos(40) = 
0.766 ≈ 
3.06 ≈ x
x ≈ 3.06
Answer:
=−7
Step-by-step explanation:
Answer:
X=8
Step-by-step explanation:
Since it has a tangent and a radius you could use the Pythagorean Theorem to find your x. This is what it would look like: