Answer:
(f⁻¹)'(b) = 1/f'(f⁻¹(b)) = 1/f'(a)
Step-by-step explanation:
The function f⁻¹(x) is the reflection of the function f(x) across the line y=x. Every point (a, b) that is on the graph of f(x) is reflected to be a point (b, a) on the graph of f⁻¹(x).
Any line with slope m reflected across the line y=x will have slope 1/m. (x and y are interchanged, so m=∆y/∆x becomes ∆x/∆y=1/m) Since f'(x) is the slope of the tangent line at (x, f(x)), 1/f'(x) will be the slope of the tangent line at (f(x), x).
Replacing x with f⁻¹(x) in the above relation, you get ...
... (f⁻¹)'(x) = 1/f'(f⁻¹(x)) will be the slope at (x, f⁻¹(x))
Putting your given values in this relation, you get
... (f⁻¹)'(b) = 1/f'(f⁻¹(b)) = 1/f'(a)
(3^3)* (3^-3)= 3^0
Remember that if you have the same base, add the exponents.
3^0=1
5(1)=5
Final answer: 5
Answer: 19 coins
Step-by-step explanation:
38-19=19
To check, take any one of the nubers and replace it with x, and when you solve for x. you'll get that number back. For example:
Let x represent the amount used to buy a notebook
19+x=38
x=38-19
x=19
Answer:
i think a is the right answer