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borishaifa [10]
3 years ago
15

Calc BC Problem. No random answers plz

Mathematics
1 answer:
stepladder [879]3 years ago
7 0

Answer:

Part A)

f(1)=2, \; f^{-1}(1)=0, \; f^\prime(1)=1.4, \; (f^{-1})^\prime(1)=\frac{10}{7}

Part B)

y=\frac{5}{14}x+\frac{4}{7}

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:

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)=\frac{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)=\frac{1}{f^\prime(0)}

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

(f^{-1})^\prime(1)=\frac{1}{0.7}=\frac{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)=\frac{1}{f^\prime(2)}

Evaluate:

(f^{-1})^\prime(4)=\frac{1}{f^\prime(2)}=\frac{1}{2.8}=\frac{10}{28}=\frac{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=\frac{5}{14}(x-4)

Solve for y:

y-2=\frac{5}{14}x-\frac{20}{14}

Adding 2 to both sides yields:

y=\frac{5}{14}x-\frac{20}{14}+\frac{28}{14}

Hence, our equation is:

y=\frac{5}{14}x+\frac{4}{7}

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Full Question:

The data show systolic and diastolic blood pressure of certain people. Find the regression equation, letting the first variable be the independent (x) variable. Find the best predicted diastolic pressure for a person with a systolic reading of 113. use a significance level of 0.05.

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Diastolic| 88 96 106 80 98 63 95 64

a. What is the regression equation?

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b. What is the best predicted value?

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Step-by-step explanation:

A. To find the regression equation here, we apply the formulas and then apply it to find the value of y given value of x:

calculate xbar and ybar which is the average of the variables:

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