0 Review problems for Test 2 (Sections R.8, 1.1-1.9) · (RS, or P9) Modeling equa
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0 Review problems for Test 2 (Sections R.8, 1.1-1.9) · (RS, or P9) Modeling equations a) A new car worth $24,000 is depreciasing in valoe by $3,000 per year. Write a formvala that models the car's value, y, in dellars, aher x years. Use the formala to determine b) After a 80% reductim, you purchase a new clodes dryer on sale for $108.. what was c) The selling price of a refrigerator is S601.70. If the markup is 10% of the dealer's cost d) The area of a rectangular wall of a barn is 200 square feet. Its length is 10 feet longer e) Each side of a square is lengthened by 3 inches. The area of this new, larger square is fter how many years the car's value will be $9,000. (y-24,000-3,000x. 5) the original price of the clothes dryer? ($540) what is the dealer's cost of the refrigerstor? (3547) than the width. Find the length and width of the wall of the barn. (10, 20) 16 square inches. Find the length of a side of the original square. (1) 2. (1.1, 1.2. )Basic concepts of functions: domain and range, function's expression, difference, difference quotient a) Find the domain of each function. x) v-7. x) 2x2+8x + 8, x)-2 b) Evaluate the function, or find the function's expression Given f(x)=2x2 +mi , find f(-1), f(-x)f(x + h) c) Find the difference and difference quotient f(x) =-3x2 + 5x + 10 , Given find f(x + h)-f(x), . Given)-find f)-f). DQ- Given f(x)- V2x, find f(x + h)-fx), DQ ) Basics of functions' graphs: point-plotting, graph piecewise function, intervals of increasing/decreasing/constant, domain, range. a) Graph cach function using the point plotting method. 3. f(x) = x2 , f(x)-f , f(x) = x*. f(x) = ! , f(x) = 1 b) Graph each piecewise function. Find the domain and range of the function from the graph -2$152 x>1 ; f(x) =13x + 2, x> 2Explanation / Answer
2)
a)
sqrt(x-7) :
We know that anything within sqrt >= 0
x - 7 >= 0
x >= 7
So, domain is [7 , inf)
b)
2x^2 + 8x + 8
This is a polynomial
So, domain is all reals
c)
2/(x-3)
Denominator cannot be 0
x - 3 = 0
x = 3
So, domain is ann numbers except 3
So, (-inf , 3) U (3 , inf) ---> ANS
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2b)
f(-1) = 2(-1)^2 + sqrt(2(-1) + 3)
= 2(1) + sqrt(-2 + 3)
= 2 + sqrt(1)
= 2 + 1
= 3 ----> ANS
f(-x) :
2(-x)^2 + sqrt(2(-x) + 3)
2x^2 + sqrt(-2x + 3)
f(-x) = 2x^2 + sqrt(3 - 2x) ---> ANS
f(x + h) :
2(x + h)^2 + sqrt(2(x+h) + 3)
2(x^2 + 2xh+ h^2) + sqrt(2x + 2h + 3)
2x^2 + 4xh + 2h^2 + sqrt(2x + 2h + 3) ---> ANS
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2c)
f(x) = -3x^2+ 5x + 10
f(x + h) = -3(x + h)^2 + 5(x + h) + 10
= -3x^2 - 3h^2 - 6xh + 5x + 5h + 10
Now, f(x+h) - f(x) becomes :
-3x^2 - 3h^2 - 6xh + 5x + 5h + 10 - (-3x^2+ 5x + 10)
= -3x^2 - 3h^2 - 6xh + 5x + 5h + 10 + 3x^2 - 5x - 10
= -3h^2 - 6xh + 5h ---> ANS
DQ = (-3h^2 - 6xh + 5h)/h
= -3h - 6x + 5 ---> ANS
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f(x) = 4/x
f(x2) = 4/x2
f(x1) = 4/x1
DQ = (4/x2 - 4/x1) / (x2 - x1)
= 4(x1 - x2)/(x1x2) divided by (x2 - x1)
= 4(-1)/(x1x2)
= -4/(x1x2) ----> ANS
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