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borishaifa [10]
2 years ago
15

Calc BC Problem. No random answers plz

Mathematics
1 answer:
stepladder [879]2 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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the standard deviation increases

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Let x₁ , x₂, .   .   .  , x₂₃ be the actual data observed by the student

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= 2.3hr

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Then x₂₃ was 10  and has been changed to 14

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Sample mean  = \frac{x_1 +x_2 +...x_2_3}{23}

\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47

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the median is x_1_2(th) values

[\frac{n +1}{2} ]^t^h value

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=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)

The new sample standard sample deviation is calculated as

= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)

Now, when we compare (1) and (2)  the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)²  ∑ (xi - 2.47)²

Therefore , the standard deviation increases

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