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3 years ago

Helpppp meeee please I can’t fail

Mathematics

1 answer:

Mazyrski [523]3 years ago

8 0

7/12 by 4/3

Step by step:

3 1/2 * 1/6 = 7/12

8*1/6 = 8/6 = 4/3

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2 years ago

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ddd [48]

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Step-by-step explanation:

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3 years ago

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

Furkat [3]

Answer:

x = -6 and x = 4

Step-by-step explanation:

In math, the critical points of a function are the points where the derivative equals zero.

So, first we will find the derivative of the function. The derivative is:

$f'(x)=3x^2 +6x-72$

Now, we are going to make the derivative equal zero and find the answers of the equation.

$3x^2 +6x-72=0\\3(x^2 +2x-24)=0\\3(x+6)(x-4)=0\\$

So we have that the critical points are the answers to this equation:

$x+6= 0 \\x= - 6$

and

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Thus, the critical points are x=-6 and x=4

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3 years ago

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2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

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3 years ago