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: c = -5
Step-by-step explanation:
Combine like terms
Subtract 4c from both sides.
Subtract 14 from both sides.
Divide by 2 to isolate c
Check:
Answer:
40
Step-by-step explanation:
28+12= 40
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Answer: 2
Step-by-step explanation:
Answer:
n = 11/3
Step-by-step explanation:
n = certain number
6n - 1 = 21
6n = 22
n = 22/6 or 11/3
