Answer:
Part A)
[tex]f(1)=2, \; f^{-1}(1)=0, \; f^\prime(1)=1.4, \; (f^{-1})^\prime(1)=\frac{10}{7}[/tex]
Part B)
[tex]y=\frac{5}{14}x+\frac{4}{7}[/tex]
Step-by-step explanation:
Please refer to the table of values.
Part A)
A. 1)
We want to find f(1).
According to the table, when x=1, f(x)=2.
Hence, f(1)=2.
A. 2)
We want to find f⁻¹(1).
Notice that when x=0, f(x)=1.
So, f(0)=1.
Then by definition of inverses, f⁻¹(1)=0.
A. 3)
We want to find f’(1).
According to the table, when x=1, f’(x)=1.4.
Hence, f’(1)=1.4.
A. 4)
We will need to do some calculus.
Let g(x) equal to f⁻¹(x). Then by the definition of inverses:
[tex]f(g(x))=x[/tex]
Take the derivative of both sides with respect to x. On the left, this will require the chain rule. Therefore:
[tex]f^\prime(g(x))\cdot g^\prime(x)=1[/tex]
Solve for g’(x):
[tex]g^\prime(x)=\frac{1}{f^\prime(g(x))}[/tex]
Substituting back f⁻¹(x) for g(x) yields:
[tex](f^{-1})^\prime(x)=\frac{1}{f^\prime({f^{-1}(x)})}[/tex]
Therefore:
[tex](f^{-1})^\prime(1)=\frac{1}{f^\prime({f^{-1}(1)})}[/tex]
We already determined previously that f⁻¹(1) is 0. Therefore:
[tex](f^{-1})^\prime(1)=\frac{1}{f^\prime(0)}[/tex]
According to the table, f’(0) is 0.7. So:
[tex](f^{-1})^\prime(1)=\frac{1}{0.7}=\frac{10}{7}[/tex]
Hence, (f⁻¹)’(1)=10/7.
Part B)
We want to find the equation of the tangent line of y=f⁻¹(x) at x=4.
First, let’s determine the points. Since f(2)=4, this means that f⁻¹(4)=2.
Hence, our point is (4, 2).
We will now need to find our slope. This will be the derivative at x=4. Therefore:
[tex](f^{-1})^\prime(4)=\frac{1}{f^\prime({f^{-1}(4)})}[/tex]
We know that f⁻¹(4)=2. So:
[tex](f^{-1})^\prime(4)=\frac{1}{f^\prime(2)}[/tex]
Evaluate:
[tex](f^{-1})^\prime(4)=\frac{1}{f^\prime(2)}=\frac{1}{2.8}=\frac{10}{28}=\frac{5}{14}[/tex]
Now, we can use the point slope form. Our point is (4, 2) and our slope at that point is 5/14.
So:
[tex]y-2=\frac{5}{14}(x-4)[/tex]
Solve for y:
[tex]y-2=\frac{5}{14}x-\frac{20}{14}[/tex]
Adding 2 to both sides yields:
[tex]y=\frac{5}{14}x-\frac{20}{14}+\frac{28}{14}[/tex]
Hence, our equation is:
[tex]y=\frac{5}{14}x+\frac{4}{7}[/tex]
