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2 years ago

Find average rate of change f(x)=x^(2)-3x-4 ; [-1,3]

1 answer:

3 0

$\bf slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\[-0.35em] \rule{34em}{0.25pt}\\\\ f(x)= x^2-3x-4\qquad \begin{cases} x_1=-1\\ x_2=3 \end{cases}\implies \cfrac{f(3)-f(-1)}{3-(-1)} \\\\\\ \cfrac{[(3)^2-3(3)-4]~~-~~[(-1)^2-3(-1)-4]}{3+1} \\\\\\ \cfrac{[-4]~~-~~[1+3-4]}{4}\implies \cfrac{[-4]-[0]}{4}\implies -1$

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2 years ago

Read 2 more answers

a 1 =− 3 1 a, start subscript, 1, end subscript, equals, minus, start fraction, 1, divided by, 3, end fraction a i = a i − 1

MrRa [10]

Answer:

$S_{75}=-506.888989761$ X 10^{32}$

Step-by-step explanation:

Given the sequence

$a_1=-\dfrac{1}{3}\\ a_i=a_{i-1}\cdot (-3)$

$a_2=a_{2-1}\cdot (-3)=a_1 \cdot (-3) = -\dfrac{1}{3}\cdot (-3)=1\\a_3=a_{3-1}\cdot (-3)=a_2 \cdot (-3) = 1 \cdot (-3)=-3\\a_4=a_{4-1}\cdot (-3)=a_3 \cdot (-3) = -3 \cdot (-3)=9\\$

Therefore the sequence is:

$-\dfrac{1}{3}, 1, -3, 9, ...$

This is a geometric sequence where the:

First Term, $a_1=-\dfrac{1}{3}$

Common ratio, r =-3

We want to determine the sum of the first 75 terms.

For a geometric sequence, the sum:

$S_n=\dfrac{a_1(1-r^n)}{1-r} \\$

Therefore:

$S_{75}=\dfrac{-\dfrac{1}{3}(1-(-3)^{75})}{1-(-3)} \\\\=\dfrac{-(1-(-3)^{75})}{4*3}\\\\S_{75}=\dfrac{(-3)^{75}-1}{12}\\\\S_{75}=-506.888989761$ X 10^{32}$

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2 years ago