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:
6
Step-by-step explanation:
Altitude = 3400t +600
21,000 = 3400t +600
21000-600= 3400t
20400/3400 = 3400t/3400
t= 6
Answered by Gauthmath
Answer:
BC=5
Step-by-step explanation:
BC^2= AB^2 + AC^2
x^2= 4^2 + 3^2
x^2= 16 +9
x^2= 25
x=
x= 5
The answer is b because the problem is starting that
