If f (2x-1) = 8x^3 - 12x^2 + 4x, what is f(5)?
Put x=3 so you get f(6–1)=f(5) hencef(5)=8*27–12*9 +4*3 =216–108+4*3 = 108+4*3=120
For Algebra 1 homework, If f(x) = 2x - 6, find f(2)?
Since you're replacing the x in f(x) with 2, then you do the same to other x. So f(2) = 2 (2) - 6 Now you just simplify and that's your answer.
What will be the remainder when f(x) = 2x^4 - 6x^3 + 2x^2-x+2 is divided by g(x) =x+2?
Given, f(x) = 2x^4 -6x^3 +2x^2 -x+2 & g(x) = x+2.Now, suppose, f(x) = g(x)×q(x)+r(x) [Remainder Theorem.]; where, q(x) is the respective quotient when f(x) is divided by g(x) & r(x) is the corresponding remainder.Putting x = -2, we get,f(-2) = g(-2)×q(-2)+r(-2)=> 2(-2)^4 -6(-2)^3 +2(-2)^2 -(-2)+2 = 0+r(-2) [since, g(-2) = 0.]=> r(-2) = 32+48+8+4i.e., r(-2) = 92.Hence, the required remainder is 92. (Ans)Hope, it'll help..!!P.S.- Feel free to comment if you have any query with it.Thank You!
If α & β are the zeros of f(x) = 2x^2 + 6x -6, then α+β= αβ, α+β > αβ, α+β < αβ, and/or α+β+αβ=0?
Of α & β are two zeros of f(x)=ax²+bx+c then ,α + β = -b/aαβ=c/aSo for your problema=2 b=6 c=-6So α + β = -6/2 = -3αβ= -6/2 = -3Hence α + β= αβ & therefore α + β - αβ = 0.
If [math]f(x) = 2x + 2[/math] , then what is [math]f(f(3))[/math] equal to?
:-DThx for taking me back to the school, buddy.How do you solve it, is the more important question, I think.Should you first calculate f(3) then use that value for f(f(3))?OrShould you get a formula for f(f(x)) and apply that on f(f(3))?That would depend upon the exam that you are writing. If it is one off, then just get f(3) = 8 and f(f(3)) = f(8) = 18.Otherwise, get f(f(x)) = f(2x+2) = 2(2x+2) + 2 = 4x+6 and apply it on multiple questions which are asked over this formula.
If f(x) = 2x and f(x+1) + f(x+3) + f(x+5) = k f(x), what is the value of k?
This is not the easiest way. I was just messing around.[math]f(x)=2x[/math][math]f(x+1)+f(x+3)+f(x+5)= 2(x+1)+2(x+1+2)+2(x+1+4)[/math][math]=2(x+1)+ 2(x+1)+4 + 2(x+1)+8 = 3[2(x+1)]+12= 6(x+1)+12=k*f(x)[/math]so, k = [math]\dfrac{6(x+1)+12}{2x}[/math] = [math]\dfrac{3(x+1)}{x} + \dfrac{6}{x}[/math]= [math]\dfrac{3x+9}{x}= 3+\dfrac{9}{x}[/math]
How do you find an expression for a cubic function f if f(5) = 300 and f(−5) = f(0) = f(6) = 0?
Since [math]f(-5) = f(0) = f(6) = 0[/math]. Then -5, 0, and 6 are roots of the function. Since the function is cubic. It is [math]f(x) = ax(x+5)(x-6)[/math]Since [math]f(5) = 300[/math] [math]a5(5+5)(5-6) = 300[/math] [math]a = -6[/math]The function therefore is [math]f(x) = -6x(x+5)(x-6)[/math]
If f(x) = 6x^2, what is the value of {f (a + b) - f (a - b)} ÷ ab?
I am surprised to see this question here.I think that it is not a right place to do post your homework.Given f(x)=6x^2=>f(a+b) =6(a+b)^2=6(a^2+b^2+2ab) f(a-b) =6(a-b)^2=6(a^2+b^2-2ab)=>f(a+b)-f(a+b)=24ab=>[f(a+b)-f(a-b)]/(ab) =24ab/ab =24.
If F(x)=3x-5, what is the value of F (2x+1)?
Hi Ahmed. You can treat x and 2x+1 just as though they are real numbers.So f(2x+1) = 3(2x+1) - 5 = 6x + 3 - 5 = 6x - 2In the above, I made use of something called the distributive law which says that if a,b and c are real numbers, then a(b+c) = ab + ac.
F(x)=2x^n+ax^2-6 is divided by (x-1), the remainder is -7. What is the value of a and n? How do you write the polynomial function completely?
We are “told” that [math]\frac{f(x)}{x-1}=g(x)-\frac{7}{x-1}[/math] for some [math]a[/math] and [math]n,[/math] and we are to find them. Multiplying both sides by [math]x-1[/math] gives [math]f(x)=(x-1)g(x)-7[/math] and hence [math]f(1)=-7[/math]. This is the “remainder theorem” mentioned in prior answer. This enables you to find [math]a[/math] but gives no information about [math]n[/math].However, what if you did not know if any [math]a[/math] and [math]n[/math] existed satisfying the condition in the statement of the problem? We might be able to find [math]a[/math] as above but we still don’t know how many choices of [math]n[/math] there are, if any, that satisfy the problem condition. How would you solve the problem fully?Using trial and error until you see a pattern is a good approach. Here are the results performing the long division [math]x-1 )\overline{ 2x^n+ax^2-6}[/math] explicitly for various [math]n[/math]:[math]n=0 \implies 2x^0+ax^2-6 = (x-1)[ax+a] + a-4[/math][math]n=1 \implies 2x^1+ax^2-6 = (x-1)[ax+(a+2)] + a-4[/math][math]n=2 \implies 2x^2+ax^2-6 = (x-1)[(a+2)x+(a+2)] + a-4[/math][math]n=3 \implies 2x^3+ax^2-6 = (x-1)[2x^2+(a+2)x+(a+2)] + a-4[/math][math]n=4 \implies 2x^4+ax^2-6 = (x-1)[2x^3+2x^2+(a+2)x+(a+2)]+a-4[/math]Clearly, choosing [math]a=-3[/math] gives the remainder [math]a-4=-7[/math] and this works for all [math]n=0,1,2,3,4.[/math] So the problem is solved for [math]n=0,1,2,3,4.[/math]By induction, you can show it holds for all larger [math]n.[/math]Inductive hypothesis: assume for some [math]n\geq 4[/math] that[math]2x^n+ax^2-6 = (x-1)[2x^{n-1}+...+2x^2+(a+2)x+(a+2)]+a-4[/math]Next, add [math]2x^{n+1}-2x^n=(x-1)2x^n[/math] to both sides and simplify to get:[math]2x^{n+1}+ax^2-6 = (x-1)[2x^n+...+2x^2+(a+2)x+(a+2)]+a-4[/math]Hence, the expression holds for [math]n+1[/math] as well so by induction this expression holds for all [math]n\geq 0.[/math]I didn’t try [math]n<0[/math]. :)