Given f(x)=(x+2)/(x-3) find f^-1(x).
To find f^-1(x) we need to set f(x)=y , y =(x+2)/(x-3). then we flip x and y . we have x =(y+2)/(y-3). multiplying both sides by x(y-3)=y+2. xy -3x=y+2. on simplifying we have y(x-1)=3x+2.The required inverse y =(3x+2)/(x-1)
What is the slope of the tangent line of the function f(x)=2x^3−4x^2−3x at x=2x=2?
To find the slope first we need to find the derivative f'(x)=6x^2-8x-3 . Now we need to plug x=2 wherever we see x in f'(x). f'(2)=6(2)^2-8(2)-3=24-16-3=24-19=5.Hence the required slope is 5.
The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended?
I would have set this up by picking a variable for one of the groups (say, "c" for "children") and then use "(total) less (what I've already accounted for)" (in this case, "2200 – c") for the other group. Using a system of equations, however, allows me to use two different variables for the two different unknowns. number of adults: a number of children: c With these variables, I can create equations for the totals they've given me: total number: a + c = 2200 total income: 4a + 1.5c = 5050 Now I can solve the system for the number of adults and the number of children. I will solve the first equation for one of the variables, and then substitute the result into the other equation: a = 2200 – c 4(2200 – c) + 1.5c = 5050 8800 – 4c + 1.5c = 5050 8800 – 2.5c = 5050 –2.5c = –3750 c = 1500 Now I can back-solve for the value of the other variable: a = 2200 – (1500) = 700 1500 children and 700 adults.