Prove that f(5)*f(6)*f(7)*f(8)*f(9)*f(10)=7 if f(x)=(x^2+x-12)/(x^2-16).

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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f(x) = (x^2 + x- 12) / (x^2 -16)

First, let us simplify the function:

We know that:

(x^2 + x -12) = (x+4)(x-3)

(x^2 -16) = (x+4)(x-4)

Noe let us substitute:

==> f(x) = (x+4)(x-3)/(x+4)(x-4)

Now reduce similar:

==> f(x) = (x-3)/(x-4)

Now let us find the values for f(5), f(6) ...f(10):

f(5) = (5-3)/(5-4) = 2/1

f(6) = (6-3)/(6-4) = 3/2

f(7) = (7-3)/(7-2) = 4/3

.....

f(n) = (n-3)/(n-4)

==>  P = f(5)*f(6)*(f(7)*f(8)*f(9)*f(10)

            = 2/1 * 3/2 * 4/3 * 5/4 * 6/5 * 7/6

             = 7.

==> P = 7

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(x) = (x^2+x-12)/(x^2-16)

To find f(5)*f(6)*f(7)*f(8)*f(9)*f(10).

We simplify (x^2+x-12)/(x^2-16) by factorising numerator and denominator.

x^2+x-12  = (x+4)(x-3)

x^2-16 = (x+4)(x-4).

Therefore f(x) = (x+4)(x-3)/(x+4)(x-4) =  (x-3)/(x-4).

Therefore f(5)*f(6)*f(7)*f(8)*f(9)*f(10) = [(5-3)/(5-4)][(6-3)/(6-4)][(7-3)/(7-4)][(8-3)/(8-4)][(9-3)/(9-4)][(10-3)/(10-4)].

f(5)*f(6)*f(7)*f(8)*f(9)*f(10) = (2*3*4*5*6*7)/(1*2*3*4*5*6).

we divide both numerator and denominator on the right side by 2*3*4*5*6 and we get:

f(5)*f(6)*f(7)*f(8)*f(9)*f(10) = 7.

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Before calculating the product f(5)*f(6)*f(7)*f(8)*f(9)*f(10)=7, we'll try to simplify the ratio (x^2+x-12)/(x^2-16).

We notice that the denominator is a difference of squares so we'll re-write it as a product:

a^2 - b^2 = (a-b)(a+b)

We'll put a^2 = x^2 and b^2 = 16

x^2 - 16 = (x-4)(x+4)

Now, we'll try to factorize the numerator. For this reason, we'll compute the roots of the expression x^2+x-12.

x^2+x-12 = 0

We'll apply the quadratic formula:

x1 = [-1 + sqrt(1 + 48)]/2

x1 = (-1+7)/2

x1 = 3

x2 = (-1-7)/2

x2 = -4

Now, we'll factorize the expression:

x^2+x-12 = (x-x1)(x-x2)

We'll substitute x1 and x2:

x^2+x-12 = (x-3)(x+4)

Now, we'll re-write f(x):

f(x) = (x-3)(x+4)/(x-4)(x+4)

We'll simplify and we'll have:

f(x) = (x-3)/(x-4)

Now, we'll plug in values for x:

x = 5 => f(5) = (5-3)/(5-4)

              f(5) = 2/1

x = 6 => f(6) = (6-3)/(6-4)

              f(6) = 3/2

x = 7 => f(7) = (7-3)/(7-4)

              f(7) = 4/3

x = 8 => f(8) = (8-3)/(8-4)

              f(8) = 5/4

x = 9 => f(9) = (9-3)/(9-4)

              f(9) = 6/5

x = 10 => f(10) = (10-3)/(10-4)

              f(10) = 7/6

Now, we'll calculate the product:

f(5)*f(6)*f(7)*f(8)*f(9)*f(10)=(2/1)(3/2)(4/3)(5/4)(6/5)(7/6)

We'll simplify like terms and we'll get:

f(5)*f(6)*f(7)*f(8)*f(9)*f(10) = 7/1

f(5)*f(6)*f(7)*f(8)*f(9)*f(10) = 7

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