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Talja [164]
4 years ago
13

A function is given. Determine the average rate of change of the function between the given values of the variable: f(x)= 4x^2;

x=3, x=3+h
Mathematics
1 answer:
nevsk [136]4 years ago
8 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}\\\\
-------------------------------

\bf f(x)= 4x^2  \qquad 
\begin{cases}
x_1=3\\
x_2=3+h
\end{cases}\implies \cfrac{f(3+h)-f(3)}{(3+h)~-~(3)}
\\\\\\
\cfrac{[4(3+h)^2]~~-~~[4(3)^2]}{\underline{3}+h-\underline{3}}\implies \cfrac{4(3^2+6h+h^2)~~-~~4(9)}{h}
\\\\\\
\cfrac{4(9+6h+h^2)~~-~~36}{h}\implies \cfrac{\underline{36}+24h+4h^2~~\underline{-~~36}}{h}
\\\\\\
\cfrac{24h+4h^2}{h}\implies \cfrac{\underline{h}(24+4h)}{\underline{h}}\implies 24+4h
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Answer:

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The amount invested in the account that paid 10%  was $6,000

The amount invested in the account that paid 12%  was $26,000

Step-by-step explanation:

Let

3x -----> the amount invested in the account that paid 8%

x -----> the amount invested in the account that paid 10%

$50,000-4x ----> the amount invested in the account that paid 12%

in this problem we have

8\%=8/100=0.08\\10\%=10/100=0.10\\12\%=12/100=0.12

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3x(0.08)+x(0.10)+(50,000-4x)(0.12)=5,160

Solve for x

0.24x+0.10x+6,000-0.48x=5,160

0.48x-0.24x-0.10x=6,000-5,160

0.14x=840

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3x=\$18,000

\$50,000-4x=\$26,000

therefore

The amount invested in the account that paid 8%  was $18,000

The amount invested in the account that paid 10%  was $6,000

The amount invested in the account that paid 12%  was $26,000

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