Question

Calc BC Problem. No random answers plz

Answer 1

Part A)

`f(1)=2, \; f^(-1)(1)=0, \; f^\prime(1)=1.4, \; (f^(-1))^\prime(1)=(10)/(7)`

Part B)

`y=(5)/(14)x+(4)/(7)`

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:

`f(g(x))=x`

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

`f^\prime(g(x))\cdot g^\prime(x)=1`

Solve for g’(x):

`g^\prime(x)=(1)/(f^\prime(g(x)))`

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

`(f^(-1))^\prime(x)=\frac{1}{f^\prime({f^(-1)(x)})}`

Therefore:

`(f^(-1))^\prime(1)=\frac{1}{f^\prime({f^(-1)(1)})}`

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

`(f^(-1))^\prime(1)=(1)/(f^\prime(0))`

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

`(f^(-1))^\prime(1)=(1)/(0.7)=(10)/(7)`

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:

`(f^(-1))^\prime(4)=\frac{1}{f^\prime({f^(-1)(4)})}`

We know that f⁻¹(4)=2. So:

`(f^(-1))^\prime(4)=(1)/(f^\prime(2))`

Evaluate:

`(f^(-1))^\prime(4)=(1)/(f^\prime(2))=(1)/(2.8)=(10)/(28)=(5)/(14)`

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

So:

`y-2=(5)/(14)(x-4)`

Solve for y:

`y-2=(5)/(14)x-(20)/(14)`

Adding 2 to both sides yields:

`y=(5)/(14)x-(20)/(14)+(28)/(14)`

Hence, our equation is:

`y=(5)/(14)x+(4)/(7)`