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butalik [34]
2 years ago
11

1. (5pts) Find the derivatives of the function using the definition of derivative.

Mathematics
1 answer:
andreyandreev [35.5K]2 years ago
8 0

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

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Answer:

1 g/cm³

Step-by-step explanation:

Volume of the model:

V=1/3bh= 1/3*100*6= 300 cm³

Density= weight/volume= 300 g/300 cm³= 1 g/cm³

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3 years ago
1-What is the sum of the series? ​∑j=152j​ Enter your answer in the box.
tangare [24]

Answer:

Please see the Step-by-step explanation for the answers

Step-by-step explanation:

1)

∑\left \ {{5} \atop {j=1}} \right. 2j

The sum of series from j=1 to j=5 is:

∑ = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)

  =  2 + 4 + 6 + 8 + 10

∑ = 30

2)

This question is not given clearly so i assume the following series that will give you an idea how to solve this:

∑\left \ {{4} \atop {k=1}} \right. 2k²

The sum of series from k=1 to j=4 is:

∑ = 2(1)² + 2(2)² + 2(3)² + 2(4)²

  = 2(1) + 2(4) + 2(9) + 2(16)

  =  2 + 8 + 18 + 32

∑ = 60

∑\left \ {{4} \atop {k=1}} \right. (2k)²

∑ = (2*1)² + (2*2)² + (2*3)² + (2*4)²

  = (2)² + (4)² + (6)² + (8)²

  = 4 + 16 + 36 + 64

∑ = 120

∑\left \ {{4} \atop {k=1}} \right. (2k)²- 4

∑ = (2*1)²-4 + (2*2)²-4 + (2*3)²-4 + (2*4)²-4

  = (2)²-4 + (4)²-4 + (6)²-4 + (8)²-4

  = (4-4) + (16-4) + (36-4) + (64-4)

  = 0 + 12 + 32 + 60

∑ = 104

∑\left \ {{4} \atop {k=1}} \right. 2k²- 4

∑ = 2(1)²-4 + 2(2)²-4 + 2(3)²-4 + 2(4)²-4

  = 2(1)-4 + 2(4)-4 + 2(9)-4 + 2(16)-4

  = (2-4) + (8-4) + (18-4) + (32-4)

  = -2 + 4 + 14 + 28

∑ = 44

3)

∑\left \ {{6} \atop {k=3}} \right. (2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)  

  = (6-10) + (8-10) + (10-10) + (12-10)

  = -4 + -2 + 0 + 2  

∑ = -4

4)

1+1/2+1/4+1/8+1/16+1/32+1/64

This is a geometric sequence where first term is 1 and the common ratio is 1/2 So

a = 1

This can be derived as

1/2/1 = 1/2 * 1 = 1/2

1/4/1/2 = 1/4 * 2/1 = 1/2

1/8/1/4 = 1/8 * 4/1  = 1/2

1/16/1/8 = 1/16 * 8/1  = 1/2

1/32/1/16 = 1/32 * 16/1  = 1/2

1/64/1/32 = 1/64 * 32/1  = 1/2

Hence the common ratio is r = 1/2

So n-th term is:

ar^{n-1} = 1(\frac{1}{2})^{n-1}

So the answer that represents the series in sigma notation is:

∑\left \ {{7} \atop {j=1}} \right. (\frac{1}{2})^{j-1}

5)

−3+(−1)+1+3+5

This is an arithmetic sequence where the first term is -3 and the common difference is 2. So  

a = 1

This can be derived as

-1 - (-3) = -1 + 3 = 2

1 - (-1) = 1 + 1 = 2

3 - 1 = 2

5 - 3 = 2

Hence the common difference d = 2

The nth term is:

a + (n - 1) d

= -3 + (n−1)2

= -3 + 2(n−1)

= -3 + 2n - 2

= 2n - 5

So the answer that represents the series in sigma notation is:

∑\left \ {{5} \atop {j=1}} \right. (2j−5)

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Answer:

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Step-by-step explanation:

The time difference t between 12.30pm and 3.30pm is 3h.

Let the distance to Marat's house be s.

The equation to calculate velocity v is given by:

v = distance / time.

Now you can write an equation for the time difference t and use the two velocities and the 90 min(= 1.5h) she stays:

3 = s/12 + 1.5 + s/2.3

Solve this equation for s:

(1/12 + 10/23)* s = 1.5

s = 2.9

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