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Question 209395This question is from textbook Elementary and Intermediate Algebra
: I am having a hard time grasping the concept of functions. My problem I am trying to solve is.... h(x)= 1/2x^2 - 3x + 2
This question is from textbook Elementary and Intermediate Algebra
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your equation is:
h(x) = (1/2)*x^2 - 3x + 2
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if you set this equation equal to 0 to solve for the roots, you can multiply both sides of this equation by 2 in order to remove the fraction coefficient from the x^2.
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your original equation of:
(1/2)*x^2 - 3x + 2 = 0
becomes:
x^2 - 6x + 4 = 0
after you multiply both sides of the equation by 2.
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it does not appear that you have integer roots so we'll solve using the quadratic formula.
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i'm assuming that you know how to use the quadratic formula so i didn't include either the formula or how to solve using it.
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if you have trouble with that, send me an email and i'll answer the question for you.
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your answer is:
x = 5.24
or:
x = .76
rounded to the nearest hundredth.
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i verified the answers are correct by plugging the whole number (not the truncated version) into the equation to see that the result is 0.
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functional notation is not real different from what you are already doing.
y = x^2
and
f(x) = x^2
mean the same thing.
y = f(x) means that y is a function of x.
the function name is f.
the argument of the function is x.
if the equation is:
y = x^2, then:
you can call the function any name you want.
you can say:
y = x^2
z = x^2
k = x^2
or you can say:
f(x) = x^2
g(x) = x^2
h(x) = x^2
or you can say:
y = f(x) = x^2
z = g(x) = x^2
k = h(x) = x^2
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y = f(x) mean the same thing.
with y, you are assigning a the variable name of y to the result of the equation.
with f(x), you are assigning the function name of f to the result of the equation.
similarly:
with z, you are assigning the variable name of z to the result of the equation.
with g(x), you are assigning the function name of g to the result of the equation.
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just to make sure you understand that any variable can be assigned to any function name, you could just have easily said that:
y = h(x) = x^2
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now to the arguments.
y = f(x) = x^2 means that the argument of the function is x which means that the independent variable is x and that x is being operated on through the relationships established in the equation to get the value of the dependent variable which is represented by the variable y or the function f.
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y = f(t) = t^2 means that the argument of the function is t which means that the independent variable is t and that t is being operated on through the relationships established in the equation to get the value of the dependent variable which is represented by the variable y or the function f.
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the argument is the independent variable of the equation.
the result of the operation of the equation is assigned to the variable name or y r to the function name of f (in this case).
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back to your equation:
h(x) = (1/2)*x^2 - 3x + 2
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this is no different than saying:
y = (1/2)*x^2 - 3x + 2
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you solve it the same way.
to solve this equation, you had to solve for x to get the roots.
whether you called it h(x) = or y = made no difference.
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here's where a difference can exist.
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your equation is:
y = (1/2)*x^2 - 3x + 2
and you are asked to solve for y when x = 5
you would go into the equation and replace x with 5 and solve.
your answer would be:
y = (1/2)*(5)^2 - 3*(5) + 2
which would be:
y = -.5
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alternately, your equation is:
f(x) = (1/2)*x^2 - 3x + 2
and you asked to solve for f(5).
you would go into the equation and replace x with 5 and solve.
your answer would be:
f(5) = (1/2)*(5)^2 - 3*(5) + 2
which would be:
f(5) = -.5
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same equation, same problem, same answer.
only difference is you assigned the result to y in the first case and to f(5) in the second case.
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when you look at the answer, however, the functional notation tells you more than the variable notation.
y = -.5 doesn't tell you much more except the answer.
f(5) = -.5 tells you the answer plus what argument you used to get it.
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in the first case you had to say y = -.5 when x = 5
in the second case, all you had to say was f(5) = -.5
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functional notation provides more information about the answer than variable notation.
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there are these and other reasons that get into higher mathematics, but the key to understand functional notation is that the process of finding the answer from the equation is the same whether you use functional notation or variable notation.
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