I have a question and then later on maybe you could check my answer?
I need to find (2fg)(-1) where f(x) = 2x^2 + 5 and g(x) = -3x What I'm planning on doing is this: (-2x^2 - 5)(3x) but I don't know what to do with the 2 that was before the fg?
put the 2 infront of the whole equation
And once I do that, I multiply 2 times (-2x^2 - 5) and 2 times (3x)?
you multiply everything in the parentheses by 2.. 2(-2x^2-5) 2(3x) 2*(-2)-2*(5)*2*(3) like this...
Okay! Does the 2 become negative? It's (2)(-1)(2x^2 + 5)(-3x)
-1 is your x value
ohh right right okay one second
Okay I got -4x^2 - 10 - 6x Is that correct? I multiplied everything by 2. Do I need to multiply -4x^2 - 10 by -6x?
now you need to substitute -1 in for every x
Okay
Wait I think I made a mistake, I multiplied everything by -1, one second
So I did 2(2x^2 + 5)(-3x) 4x^2 + 10 - 6x 4(-1)^2 + 10 - 6(-1) 4 + 10 - (-6) = 20 Is that correct?
well 4* (-1)=-4 and then you would have to sqaure it so it would be -4+10*(-6)
Oh, I thought you have to do -1^2 first
-64 then?
oops yea sorry you do do the -1^2 first
oh so then it is 20?
yea sorry
np
wait did you get that answer by subtracting -6 or multiplying it?
-6 should have been multiplied
oops I should be multiplying... >< (4x^2 + 10)(-6x) (4(-1)^2 + 10)(-6(-1)) (4 + 10)(6) 14 * 6 = 84
14*(-6)=-84
no but, it's -3x times 2 which is -6x and the x is -1 so (-6)(-1) equals 6?
oh sorry im doing 2 things at once your right its 84
(2fg)(-1) where f(x) = 2x^2 + 5 and g(x) = -3x Are you sure we are not doing function composition, which is written as \( 2 f \circ g \) ? If not, then we interpret this to mean we multiply the two functions: 2 * f(x)*g(x) we could do this the long way and get 2 * (2x^2 + 5) * (-3x) = -12x^3-30x now we can find the value for any x, including x=-1 -12* (-1)^3 - 30 (-1) = 12+30= 42 the other way to do this is start with 2 * f(x)*g(x) and replace x with -1: 2 * f(-1) * g(-1) now find f(-1) = 2*(-1)^2 + 5 = 2+5= 7 and g(-1)= -3*-1= 3 now find 2 * 7 * 3= 42
That's how it was written, thanks Phi, thanks B. Sanders!
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