Answer:
J
Step-by-step explanation:
if you multiply any number by a number lower than 1, then the value lowers. J is the only thing that has both numbers greater than one
2.8.1
![f(x) = \dfrac4{\sqrt{3-x}}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cdfrac4%7B%5Csqrt%7B3-x%7D%7D)
By definition of the derivative,
![f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto0%7D%20%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D)
We have
![f(x+h) = \dfrac4{\sqrt{3-(x+h)}}](https://tex.z-dn.net/?f=f%28x%2Bh%29%20%3D%20%5Cdfrac4%7B%5Csqrt%7B3-%28x%2Bh%29%7D%7D)
and
![f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}](https://tex.z-dn.net/?f=f%28x%2Bh%29-f%28x%29%20%3D%20%5Cdfrac4%7B%5Csqrt%7B3-%28x%2Bh%29%7D%7D%20-%20%5Cdfrac4%7B%5Csqrt%7B3-x%7D%7D)
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)}}](https://tex.z-dn.net/?f=f%28x%2Bh%29-f%28x%29%20%3D%20%5Cdfrac%7B4%5Csqrt%7B3-x%7D%20-%204%5Csqrt%7B3-%28x%2Bh%29%7D%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%7D)
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)}](https://tex.z-dn.net/?f=f%28x%2Bh%29%20-%20f%28x%29%20%3D%20%5Cdfrac%7B%5Cleft%284%5Csqrt%7B3-x%7D%20-%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D%20%5C%5C%5C%5C%20f%28x%2Bh%29%20-%20f%28x%29%20%3D%20%5Cdfrac%7B%5Cleft%284%5Csqrt%7B3-x%7D%5Cright%29%5E2%20-%20%5Cleft%284%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%5E2%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D%20%5C%5C%5C%5C%20f%28x%2Bh%29%20-%20f%28x%29%20%3D%20%5Cdfrac%7B16%283-x%29%20-%2016%283-%28x%2Bh%29%29%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D%20%5C%5C%5C%5C%20f%28x%2Bh%29%20-%20f%28x%29%20%3D%20%5Cdfrac%7B16h%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D)
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}}}](https://tex.z-dn.net/?f=%5Cdfrac%7Bf%28x%2Bh%29-f%28x%29%7Dh%20%3D%20%5Cdfrac%7B16%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-%28x%2Bh%29%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-%28x%2Bh%29%7D%5Cright%29%7D%20%5C%5C%5C%5C%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto0%7D%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7Dh%20%3D%20%5Cdfrac%7B16%7D%7B%5Csqrt%7B3-x%7D%5Csqrt%7B3-x%7D%5Cleft%284%5Csqrt%7B3-x%7D%20%2B%204%5Csqrt%7B3-x%7D%5Cright%29%7D%20%5C%5C%5C%5C%20%5Cimplies%20f%27%28x%29%20%3D%20%5Cdfrac%7B16%7D%7B4%5Cleft%28%5Csqrt%7B3-x%7D%5Cright%29%5E3%7D%20%3D%20%5Cboxed%7B%5Cdfrac4%7B%283-x%29%5E%7B3%2F2%7D%7D%7D)
3.1.1.
![f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3](https://tex.z-dn.net/?f=f%28x%29%20%3D%204x%5E5%20-%20%5Cdfrac1%7B4x%5E2%7D%20%2B%20%5Csqrt%5B3%5D%7Bx%7D%20-%20%5Cpi%5E2%20%2B%2010e%5E3)
Differentiate one term at a time:
• power rule
![\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4](https://tex.z-dn.net/?f=%5Cleft%284x%5E5%5Cright%29%27%20%3D%204%5Cleft%28x%5E5%5Cright%29%27%20%3D%204%5Ccdot5x%5E4%20%3D%2020x%5E4)
![\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac1%7B4x%5E2%7D%5Cright%29%27%20%3D%20%5Cdfrac14%5Cleft%28x%5E%7B-2%7D%5Cright%29%27%20%3D%20%5Cdfrac14%5Ccdot-2x%5E%7B-3%7D%20%3D%20-%5Cdfrac1%7B2x%5E3%7D)
![\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}](https://tex.z-dn.net/?f=%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%27%20%3D%20%5Cleft%28x%5E%7B1%2F3%7D%5Cright%29%27%20%3D%20%5Cdfrac13%20x%5E%7B-2%2F3%7D%20%3D%20%5Cdfrac1%7B3x%5E%7B2%2F3%7D%7D)
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}}}](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%20%5Cboxed%7B20x%5E4%20%2B%20%5Cdfrac1%7B2x%5E3%7D%20%2B%20%5Cdfrac1%7B3x%5E%7B2%2F3%7D%7D%7D)
One hour. If they both work together, Fred is automatically gonna be almost done in 40 minutes but Vanessa is holding him back. So it then gets exchanged to 60 minutes.
Answer:
$5
Step-by-step explanation:
25-5 =20
20/4 =5
No, I don't agree with her expression because the numbers of quarters and nickels have already been given. To calculate the amount of money that Jackie has, she needs to multiply the number of coins by the value of the coin.
The expression should have been 0.25q+0.05n with q and n being the number of quarters and nickels that she has.