Calc BC Problem. No random answers plz

Part A)

Part B)

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:

Take the derivative of both sides with respect to x. On the left, this will require the chain rule. Therefore:

Solve for g’(x):

Substituting back f⁻¹(x) for g(x) yields:

Therefore:

We already determined previously that f⁻¹(1) is 0. Therefore:

According to the table, f’(0) is 0.7. So:

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:

We know that f⁻¹(4)=2. So:

Evaluate:

Now, we can use the point slope form. Our point is (4, 2) and our slope at that point is 5/14.

So:

Solve for y:

Adding 2 to both sides yields:

Hence, our equation is: