could someone work these few questions out for me?

edit: if you cant read the size the questions are i will gladly resize the image, sorry about that, i didnt know it was so small!

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- Apr 19th 2007, 02:26 PMfatcow2000Help with some simple problems (functions, factors etc)!!!
could someone work these few questions out for me?

edit: if you cant read the size the questions are i will gladly resize the image, sorry about that, i didnt know it was so small! - Apr 19th 2007, 03:03 PMJhevon
http://www.mathhelpforum.com/math-he...-etc-math-.jpg

Remeber how function notation works. whatever is in the brackets goes to a designated poistion in the formula.

example: if f(x) = x^2 + 3

once i place anything in the brackets besides x, it takes the place of x in the formula.

so f(2) = (2)^2 + 3

f(7) = (7)^2 + 3

f(4.5678) = (4.5678)^2 + 3

f(Jhevon) = (Jhevon)^2 + 3

now to your first question:

we have f(x) = 3x - 4 ........is that a minus? it looks a bit blurry

and g(x) = 6 - 4x^2

we want f(g(2)), now remember, whatever we put in the brackets will replace x in the formula for f(x).

so f(g(2)) = 3(g(2)) - 4

but what is g(2)? let's find out.

since we have a 2 in the brackets, it means we replace the x (or x's, if there are more than one, replace them all) with 2

so g(2) = 6 - 4(2)^2 = 6 - 16 = -10

so f(g(2)) = f(-10) = 3(-10) - 4 = -30 - 4 = -34

Your second question is a simultaneous equations question, there are two main ways to do a problem of this nature, by elimination and by substitution. i will show you both ways.

Elimination: This is where we try to multiply one of the equations by a number so that one variable has the same coefficient in both equations. then we add (or subtract) the two equations to ELIMINATE that variable. let's see how it works

3x + 4y = 24 ..................(1)

2x - 5y = 39 ..................(2)

6x + 8y = 48 ..................(3) = (1)*2 i multiplied equation (1) by 2

6x - 15y = 117 ...............(4) = (2)*3 i multiplied equation (2) by 3

now we see that x has a coefficient of 6 in both equations, so we can subtract one equation from the other and eliminate x, so we can solve for y. once we've found y, we plug it back into either equation to find x.

=> 23y = -69 .................(3) - (4) i subtracted (4) from (3)

=> y = -3

now plug that value for y into any of the above equations

But 3x + 4y = 24

=> 3x + 4(-3) = 24

=> 3x - 12 = 24

=> 3x = 24 + 12 = 36

=> x = 36/3 = 12

so the solution is: x = 12, y = -3

Substitution: this is where we use one equation and solve for one variable in terms of the other. then we SUBSTITUTE that expression in the other equation to find the other variable. then we substitute that value again in to any of the original equations to find the variable we originally substituted. confused? let's see it in action.

3x + 4y = 24 ..................(1)

2x - 5y = 39 ..................(2)

from (2) we see that x = (39 + 5y)/2, substitute (39 + 5y)/2 for x in equation (1), we get:

3[(39 + 5y)/2] + 4y = 24

=> (117 + 15y)/2 + 4y = 24

=> 117 + 15y + 8y = 48

=> 23y = -69

=> y = -3

But 3x + 4y = 24

=> 3x + 4(-3) = 24

=> 3x - 12 = 24

=> 3x = 24 + 12 = 36

=> x = 36/3 = 12

so the solution is: x = 12, y = -3

Obviously for this particular eqaution, the elimination method was easier. we knew it would be since we ended up with fractions in the first place.

Next Question:

Factor

c^2 - 25d^2

this is the difference of two squares, you should be able to recognize such problems easily, they show up all the time. to factor the difference of two squares, we follow the formula: x^2 - y^2 = (x + y)(x - y), so:

c^2 - 25d^2 = c^2 - (5d)^2 = (c + 5d)(c - 5d)

Next Question: 10x^2 - 39x + 14

this is one of the harder kinds of factoring for quadratics. whenever the coefficient of x^2 is not one, we do the following:

- multiply the x^2 term by the lone constant, here we get: 10x^2 * 14 = 140x^2

- take the middle term, here we get: -39x

- think of two numbers that multiplying we get the product, and adding we get the middle term. that is, two numbers that we multiply them and get 140x^2 and add and get -39x

- once those numbers are found, split the middle term into them. so you will end up with four terms

- factor out the common factor from the first two terms, then from the last two terms. then factor the common factor from both.

confused? let's see what happens

10x^2 - 39x + 14

we get:

140x^2

-39x

aha! -4x and -35x work! Yay!

so we get:

10x^2 - 35x - 4x + 14 ........5x is common to the first two, 2 is common to the second 2

=> 5x(2x - 7) - 2(2x - 7) ......now 2x - 7 is common to both, factor it out

=> (2x - 7)(5x - 2) ..............and we're done, not so bad was it?

Finally, last question (i have to leave after this)

5^2 * a^2 * b * c^5

the degree of a term is simply the sum of all the powers, so the degree of the above is:

2 + 2 + 1 + 5 = 10