What is 2/22 simplify

Use the four-step definition of the derivative to find f'(x) if f(x) = −4x^3 −1.

guapka [62]

$\stackrel{de finition \textit{ of a derivative as a limit}}{\lim\limits_{h\to 0}~\cfrac{f(x+h)-f(x)}{h}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{[-4(x+h)^3-1]~~ - ~~[-4x^3-1]}{h} \\\\\\ \cfrac{[-4(x^3+3x^2h+3xh^2+h^3)-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{[-4x^3-12x^2h-12xh^2-4h^3-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{-4x^3-12x^2h-12xh^2-4h^3-1+4x^3+1}{h}\implies \cfrac{-12x^2h-12xh^2-4h^3}{h}$

$\cfrac{h(-12x^2-12xh-4h^2)}{h}\implies -12x^2-12xh-4h^2 \\\\\\ \lim\limits_{h\to 0}~-12x^2-12xh-4h^2\implies \lim\limits_{h\to 0}~-12x^2-12x(0)-4(0)^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \lim\limits_{h\to 0}~-12x^2~\hfill$

7 0

1 year ago

Explain please. .........

timofeeve [1]

It means -1x - 2 = 3 those bars make all the number positive

4 0

3 years ago

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Find the roots of:

Elena-2011 [213]

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{1) \ 2x^3-7x^2+8x-3=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -3 are 1, -1, 3, +3. So:

| 2 -7 8 -3

1 | 2 -5 3

| 2 -5 3 0

1 | 2 -3

2 -3 0

So the factorization is (x-1)² (2x-3)=0. So:

$\bf{ x_1=x_2=1 \qquad x_2=\dfrac{3}{2} }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{2) \ x^3-x^2-4=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -4 are 1, -1, 2, -2, 4, -4. So:

| 1 -1 0 -4

2 | 2 2

1 2 2 0

So the factorization is (x-2)(x²+x+2)=0 . When calculating the discriminant of the trinomial, it is concluded that it has no roots since the result is negative. So you only have one solution.

$\bf{ 1^2-4(2)(2)=1-16=-15 < 0 \quad \Longrightarrow \quad x=2 }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{3) \ 6x^3+7x^2-9x+2=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of 2 are 1, -1, 2, -2. So:

| 6 7 9 2

-2 | -12 10 -2

6 -5 1 0

So the factorization is (x+2)(6x²-5x+1)=0 . The quadratic equation is solved by the general formula:

$\bf{ x_{2, 3}&=\dfrac{5\pm \sqrt{(5)^2-4(6)(1)}}{2(6)}=\dfrac{5\pm \sqrt{25-24}}{12}=\dfrac{5\pm 1}{12} }}$

$\large\displaystyle\text{$\begin{gathered}\sf \begin{matrix} x_1=-2&\ \ \ \ \ \ x_{2}=\dfrac{6}{12} \qquad &\ \ \ x_3=\dfrac{4}{12}\\ &\ \ \ x_2=\dfrac{1}{2} \qquad &x_3=\dfrac{1}{3} \end{matrix} \end{gathered}$}$

6 0

1 year ago

The answer to these please!

igomit [66]

$\sqrt[4]{(4a^{2})^{6}} = \sqrt[4]{4096a^{12}} = 8a^{3}$

$\sqrt[7]{(b^{\frac{5}{3}})^{8}} = \sqrt[7]{b^\frac{40}{3}} = b^{\frac{40}{21}} = \sqrt[21]{b^{40}} = \sqrt[21]{b^{21} * b^{19}} = \sqrt[21]{b^{21}} \sqrt[21]{b^{19}} = b\sqrt[21]{b^{19}}$

$\sqrt[3]{\frac{c^{9}}{c^{3}}} = \sqrt[3]{c^{6}} = c^{2}$

$\sqrt[3]{27d^{6} * 8d^{-4}} = \sqrt[3]{27d^{6} * \frac{8}{d^{4}}} = \sqrt[3]{216d^{2}} = 6\sqrt[3]{d^{2}} = 6d^{\frac{2}{3}}$

4 0

2 years ago

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A photograph 5 inches wide by 6 inches high is to be enlarged so that the new height is 20 inches. what will the width of the en

Vika [28.1K]

The new width is 16.6666666667 inches, which can be rounded to 16.7.

Width: 5 ?
Height: 6 20

We know that based on the information provided, the new height is 20 inches.

So, in order to solve, you can cross multiply (5 x 20), which would be 100. Then

divide by 6, to get 16.6666666667 or 16.7.

7 0

2 years ago