Function Problem I Got Incorrect Answers

[tsu.j.w.dev]

This is a problem that I Am Working but I got an error on these functions problems:

If f(x) = 3x² - x + 2 ---> find 1.) f(a+1), 2.) [f(a)]²

1.) f(a+1) = 3(a+1)² - (a+1) + 2
= 3[a² + 2(a)(1) + (1)²] - a - 1 + 2
= 3a² +2a + 1 -a -1 +2
= 3a² + a + 2 ------> THIS ANSWER IS INCORRECT

= 3a² + 5a + 4 -------> THIS IS THE CORRECT ANSWER

** CAN SOMEONE WORK OUT THIS PROBLEM WITH THE CORRECT SOLUTIONS STEPS **


2.) [f(a)]² = [3(a)² - a + 2)]²
= 9a⁴ - a² + 4 ---------> THIS ANSWER IS INCORRECT


= 9a⁴ - 6a² - +13 a² - 4a + 4 ----------> THIS IS THE CORRECT ANSWER

** CAN SOMEONE WORK OUT THIS PROBLEM WITH THE CORRECT SOLUTIONS STEPS **

 

[tsu.j.w.dev]

I Am Working On A Algebra Problem But Got Two Incorrect Answers

If f(x) = 3x² - x + 2 -------> Find 1.) (f( a+1 ) 2.) [f(a)]²

1.) f(a + 1) = 3(a + 1)² - (a + 1) + 2
= 3[(a² + 2(a)(1) + (1)²] - a - 1 +2
= 3a² + 2a + 1 -a -1 +2
= 3a² + a + 2 ----------------> INCORRECT ANSWER


= 3a² + a + 2 -----------------> THIS IS THE CORRECT ANSWER

** SOMEWHERE I WORKED THE PROBLEM WITH THE WRONG STEP CAN SOMEONE WORK THE CORRECT SOLUTION **

2.) [f[(a)]² = [3(a)² - a + 2)]²
= 9a⁴ - a² + 4 -------------------> INCORRECT ANSWER


= 9a⁴ - 6a²+ 13 a² - 4a + 4 -------> CORRECT ANSWER

** SOMEWHERE I WORKED THE PROBLEM WITH THE WRONG STEP CAN SOMEONE WORK THE CORRECT SOLUTION **

 

[stapel]

This is a problem that I Am Working but I got an error on these functions problems:

If f(x) = 3x² - x + 2 ---> find 1.) f(a+1), 2.) [f(a)]²

1.) f(a+1) = 3(a+1)² - (a+1) + 2
= 3[a² + 2(a)(1) + (1)²] - a - 1 + 2
= 3a² +2a + 1 -a -1 +2

How did you get that 3 times 2a was just 3a?

2.) [f(a)]² = [3(a)² - a + 2)]²
= 9a⁴ - a² + 4

Powers do not "distribute" across addition!

To learn how to multiply through parentheses, try here. To learn how to multiply polynomials, try here.

 

[Ishuda]

I Am Working On A Algebra Problem But Got Two Incorrect Answers

If f(x) = 3x² - x + 2 -------> Find 1.) (f( a+1 ) 2.) [f(a)]²

1.) f(a + 1) = 3(a + 1)² - (a + 1) + 2
= 3[(a² + 2(a)(1) + (1)²] - a - 1 +2
= 3a² + 2a + 1 -a -1 +2
= 3a² + a + 2 ----------------> INCORRECT ANSWER


= 3a² + a + 2 -----------------> THIS IS THE CORRECT ANSWER

** SOMEWHERE I WORKED THE PROBLEM WITH THE WRONG STEP CAN SOMEONE WORK THE CORRECT SOLUTION **

2.) [f[(a)]² = [3(a)² - a + 2)]²
= 9a⁴ - a² + 4 -------------------> INCORRECT ANSWER


= 9a⁴ - 6a²+ 13 a² - 4a + 4 -------> CORRECT ANSWER

** SOMEWHERE I WORKED THE PROBLEM WITH THE WRONG STEP CAN SOMEONE WORK THE CORRECT SOLUTION **

When things get 'complicated', sometimes it is easier to do terms individually. For example, lets look at f(x)=5 x² - 4 x + 1 and find f(a+1)
f(a+1) = 5 (a+1)² -4 (a+1) + 1

[FONT=courier new][FONT=fixedsys]5 (a+1)[SUP]2[/SUP] = 5 (a[SUP]2[/SUP] + 2 a + 1) = 5 a[SUP]2[/SUP] + 10 a +  5
-4(a+1)  = -4 a - 4         =      -  4 a -  4  
1                           =             +  1
                               ________________
f(a+1)                      = 5 a[SUP]2[/SUP] + 6 a  +  2 [/FONT] 
[/FONT]

You would do the same sort of thing for [f(a)]²
[f(a)]²=[5 a² - 4 a + 1]² = [5 a² - 4 a + 1] * [5 a² - 4 a + 1]

[FONT=fixedsys]5a[SUP]2[/SUP][5 a[SUP]2[/SUP] - 4 a + 1] = 25a[SUP]4[/SUP] - 20a[SUP]3[/SUP] +  5a[SUP]2[/SUP]
-4a[5 a[SUP]2[/SUP] - 4 a + 1] =      - 20a[SUP]3[/SUP] + 16a[SUP]2[/SUP] - 4a
 1 [5 a[SUP]2[/SUP] - 4 a + 1] =             +  5a[SUP]2[/SUP] - 4a + 1
                        _______________________________  
[f(a)][SUP]2[/SUP]             = 25a[SUP]4[/SUP] - 40a[SUP]3[/SUP] + 26a[SUP]2[/SUP] - 8a + 1
[/FONT]

Once one has practised this for some time, it gets much easier.

 

[pka]

This is a problem that I Am Working but I got an error on these functions problems:
If f(x) = 3x² - x + 2 ---> find 1.) f(a+1), 2.) [f(a)]²

I confess that I do not follow the other replies.

\(\displaystyle \Large (x+y+z)^2=x^2+y^2+2xy+2xz+2yz\)

\(\displaystyle \large [f(a)]^2=[3a^2-a+2]^2=(3a^2)^2+(-a)^2+(2)^2+2(3a^2)(-a)+2(3a^2)(2)+2(-a)(2)\)

 

[Harry_the_cat]

I confess that I do not follow the other replies.

\(\displaystyle \Large (x+y+z)^2=x^2+y^2+2xy+2xz+2yz\) \(\displaystyle +z^2\)

\(\displaystyle \large [f(a)]^2=[3a^2-a+2]^2=(3a^2)^2+(-a)^2+(2)^2+2(3a^2)(-a)+2(3a^2)(2)+2(-a)(2)\)

See red above.

 

[Harry_the_cat]

This is a problem that I Am Working but I got an error on these functions problems:

If f(x) = 3x² - x + 2 ---> find 1.) f(a+1), 2.) [f(a)]²

1.) f(a+1) = 3(a+1)² - (a+1) + 2
= 3[a² + 2(a)(1) + (1)²] - a - 1 + 2
= 3a² +2a + 1 -a -1 +2 No this should be \(\displaystyle 3a^2+6a+3-a-1+2\) ...You need to distribute the 3 to all term in the square brackets.
= 3a² + a + 2 ------> THIS ANSWER IS INCORRECT

= 3a² + 5a + 4 -------> THIS IS THE CORRECT ANSWER

** CAN SOMEONE WORK OUT THIS PROBLEM WITH THE CORRECT SOLUTIONS STEPS **


2.) [f(a)]² = [3(a)² - a + 2)]²
= 9a⁴ - a² + 4 ---------> THIS ANSWER IS INCORRECT Think of it as:\(\displaystyle (3a^2-a+2)(3a^2-a+2)\) and expand out to get 9 terms and then collect like terms.


= 9a⁴ - 6a² - +13 a² - 4a + 4 ----------> THIS IS THE CORRECT ANSWER
** CAN SOMEONE WORK OUT THIS PROBLEM WITH THE CORRECT SOLUTIONS STEPS **

See comments in red.