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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:

15.98

Step-by-step explanation:

∑₁¹⁰ 8(\frac{1}{4} )^{n-1}

=8[1+\frac{1}{4} +\frac{1}{16} + ........ +t_{10}]

We have to use the formula of sum of a G.P. series with common ratio r < 1.

If the first term is a, the common ratio is r and the number of terms is n then the sum [a + ar + ar² + ar³ + ......... up to n terms] is given by  

[a \times\frac{1-r^{n} }{1-r}]

= 8[1 \times \frac{1-\frac{1}{2^{10} } }{1-\frac{1}{2} } ]

= 15.98 (Answer)

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Answer:

See below

Step-by-step explanation:

Expand L side to get

4n -12   -2n  <  -10 + 2n     simplify L side

2n -12   <  -10 +2n      subtract 2n from both sides

<u>-12 < -10 </u>    <===== so the equation is true for ALL n

(-∞  ,  + ∞ )

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The cost of producing a new watch is $50. At a price of $100, watches will most likely be ____ and ____.
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7 0
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Please Please help asap!!! Thank you
natali 33 [55]

Step-by-step explanation:

well, the starting equation and the target format have been given.

let's do the multiplications and compare the target with the starting information.

from there we see what is different or missing.

x² + 14x + 13 = 0

and

(x - p)² = q

x² - 2px + p² = q

x² - 2px + (p² - q) = 0

now let's compare the different parts :

x² = x²

-2px = 14x

-2p = 14

p = -7

p² - q = 13

-7² - q = 13

49 - q = 13

36 - q = 0

q = 36

so, the square (x - p)² = (x + 7)² is completed when

x² + 14x + 49 = 0

but we have only "+ 13". so we need to add 36 to get 49. but we need to do it on both sides, to keep the equation true :

x² + 14x + 13 + 36 = 36

x² + 14x + 49 = 36

(x + 7)² = 36

just as we calculated already above.

and now this can be solved by pulling the square root on both sides (a quadratic equation has always 2 solutions)

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x2 = -6 - 7 = -13

3 0
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