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:
0=24 so theirs no solution
Step-by-step explanation:
Answer:
7*3x=18
-7 -7
3x=11
3x/3 11/3
x=3.67
Step-by-step explanation:
Answer:
1191057210
Step-by-step explanation:
Reorder the terms
Evaluate the power
Multiply the numbers
(x4−3x3+4x2−8)/(x+1) = x3−4x2<span>+8x−8.</span>
