Question

precalc lesson 1.7 homework
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1. let $f(x) = x^2 + 3x - 1$ and $g(x) = 10 - 2x$. find $f(x) - g(x)$.

1. let $f(x) = \sqrt{20 - x}$ and $g(x) = \sqrt{10x - 4}$. let $h(x) = f(x)g(x)$. find $h(4)$.

1. paul and polly sell lemonade in front of their house in the summer. during the month of june, they sold about 26 cups of lemonade a day at a price of $0.50 per cup. in july, they plan to increase their cups sold by 2 cups every day with some clever new advertising. they will also decrease the price per cup by 1 cent every day. write an equation for a function $l(d)$ that gives the amount of money paul and polly make selling lemonade on day $d$ in july.

Explanation:

Problem 1

Step1: Substitute the functions

We know \( f(x)=x^{2}+3x - 1 \) and \( g(x)=10 - 2x \). So \( f(x)-g(x)=(x^{2}+3x - 1)-(10 - 2x) \).

Step2: Distribute the negative sign

\( (x^{2}+3x - 1)-(10 - 2x)=x^{2}+3x - 1-10 + 2x \).

Step3: Combine like terms

Combine the \( x \) terms and the constant terms: \( x^{2}+(3x + 2x)+(-1-10)=x^{2}+5x-11 \).

Step1: Find \( h(x) \) formula

Given \( h(x)=f(x)g(x) \), \( f(x)=\sqrt{20 - x} \) and \( g(x)=\sqrt{10x - 4} \), so \( h(x)=\sqrt{20 - x}\cdot\sqrt{10x - 4} \). Using the property \( \sqrt{a}\cdot\sqrt{b}=\sqrt{ab} \) (for \( a\geq0,b\geq0 \)), we get \( h(x)=\sqrt{(20 - x)(10x - 4)} \).

Step2: Substitute \( x = 4 \)

Substitute \( x = 4 \) into \( h(x) \): \( h(4)=\sqrt{(20 - 4)(10\times4 - 4)}=\sqrt{16\times(40 - 4)}=\sqrt{16\times36} \).

Step3: Simplify the square root

We know that \( \sqrt{16\times36}=\sqrt{16}\times\sqrt{36}=4\times6 = 24 \).

Step1: Define variables for cups sold

In June, they sold 26 cups per day. In July, they increase by 2 cups every day. So on day \( d \) in July, the number of cups sold \( C(d)=26 + 2d \) (assuming day 1 of July is \( d = 1 \), starting from 26 cups and increasing by 2 each day).

Step2: Define variables for price per cup

In June, the price is \( \$0.50=50 \) cents. In July, they decrease by 1 cent every day. So on day \( d \) in July, the price per cup \( P(d)=50 - d \) cents (or \( \$(0.50-0.01d) \)).

Step3: Find the revenue function

Revenue \( L(d) \) is (number of cups sold) times (price per cup). So \( L(d)=(26 + 2d)(0.50-0.01d) \). We can expand this:
First, multiply using the distributive property (FOIL method):
\( (26)(0.50)+26(-0.01d)+2d(0.50)+2d(-0.01d) \)
\( =13-0.26d + d-0.02d^{2} \)
Combine like terms: \( L(d)=-0.02d^{2}+( - 0.26d + d)+13=-0.02d^{2}+0.74d + 13 \) (in dollars) or in cents \( L(d)=(26 + 2d)(50 - d)=1300-26d + 100d-2d^{2}=-2d^{2}+74d + 1300 \) cents.

Answer:

\( f(x)-g(x)=x^{2}+5x - 11 \)

Problem 2