vgamelover123
New member
- Joined
- Jun 6, 2020
- Messages
- 16
My work. Why does it = 353/27?
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
What am I plugging in wrong for find the average rate of change? g(x)=?
- Thread starter vgamelover123
- Start date
D
Deleted member 4993
Guest
g(-1) = 6*(-1)2 + \(\displaystyle \frac{4}{-1^3} \ \)My work. Why does it = 353/27?
View attachment 19635
= 6 * (-1) * (-1) + \(\displaystyle \frac{4}{(-1) * (-1) * (-1)} \ \)
= (6) + \(\displaystyle \frac{4}{(-1)} \ \)
= 6 - 4
= 2 (positive number)
There are more mistakes. Work all the way with fractions - since the required answer is in fraction. Do not convert to decimals.
Please check your arithmetic - carefully.
None of what you wrote down makes any sense.
[MATH]g(x) = 6x^2 + \dfrac{4}{x^3} = \dfrac{6x^5 + 4}{x^3} \implies \\ g(-1) = \dfrac{6(-1)^5 + 4}{(-1)^3} = \dfrac{- 6 + 4}{- 1} = \dfrac{-2}{-1} = 2 \text { and } \\ g(3) = \dfrac{6(3^5) + 4}{3^3} = \dfrac{6 * 243 + 4}{27} = \dfrac{1458 + 4}{27} = \dfrac{1462}{27}.\\ \therefore \dfrac{2 - \dfrac{1462}{27}}{-1 - 3} = \dfrac{- \dfrac{1408}{27}}{-4} = \dfrac{1408}{27} * \dfrac{1}{4} = \dfrac{352}{27} [/MATH]
[MATH]g(x) = 6x^2 + \dfrac{4}{x^3} = \dfrac{6x^5 + 4}{x^3} \implies \\ g(-1) = \dfrac{6(-1)^5 + 4}{(-1)^3} = \dfrac{- 6 + 4}{- 1} = \dfrac{-2}{-1} = 2 \text { and } \\ g(3) = \dfrac{6(3^5) + 4}{3^3} = \dfrac{6 * 243 + 4}{27} = \dfrac{1458 + 4}{27} = \dfrac{1462}{27}.\\ \therefore \dfrac{2 - \dfrac{1462}{27}}{-1 - 3} = \dfrac{- \dfrac{1408}{27}}{-4} = \dfrac{1408}{27} * \dfrac{1}{4} = \dfrac{352}{27} [/MATH]
Last edited:
vgamelover123
New member
- Joined
- Jun 6, 2020
- Messages
- 16
But why raise it to the power of 5?None of what you wrote down makes any sense.
[MATH]g(x) = 6x^2 + \dfrac{4}{x^3} = \dfrac{6x^5 + 4}{x^3} \implies \\ g(-1) = \dfrac{6(-1)^5 + 4}{(-1)^3} = \dfrac{- 6 + 4}{- 1} = \dfrac{-2}{-1} = 2 \text { and } \\ g(3) = \dfrac{6(3^5) + 4}{3^3} = \dfrac{6 * 243 + 4}{27} = \dfrac{1458 + 4}{27} = \dfrac{1462}{27}.\\ \therefore \dfrac{2 - \dfrac{1462}{27}}{-1 - 3} = \dfrac{- \dfrac{1408}{27}}{-4} = \dfrac{1408}{27} * \dfrac{1}{4} = \dfrac{352}{27} [/MATH]
D
Deleted member 4993
Guest
x2 * x3 = x(2+3)But why raise it to the power of 5?
I don't like adding numerical fractions because I make mistakes. So I did it algebraically to simplify my arithmetic.But why raise it to the power of 5?
[MATH]g(x) = 6x^2 + \dfrac{4}{x^3} = \dfrac{6x^2 * x^3}{x^3} + \dfrac{4}{x^3} = \dfrac{6x^{2+3} + 4}{x^3} = \dfrac{6x^5 + 4}{x^3}.[/MATH]
It is just a slightly different way to state the function. But your problem was (a) multiple errors in arithmetic and (b) apparent lack of understanding of function notation. If you are starting calculus, you need a rapid but deep review of functions because 95% of the course involves working with them.