mathShow 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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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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First, we'll simplify the fraction (x^2+x-12)/(x^2-16).

We notice that the denominator is a difference of squares:

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

Let a^2 = x^2 and b^2 = 16

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

We'll factorize the numerator.

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 reduce and we'll get:

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 reduce 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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