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a) f(x) = 6x - 5 and g(x) = \frac{x + 5}{6} - Leaving Cert Mathematics - Question 3 - 2020

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a)--f(x)-=-6x---5-and-g(x)-=-\frac{x-+-5}{6}-Leaving Cert Mathematics-Question 3-2020.png

a) f(x) = 6x - 5 and g(x) = \frac{x + 5}{6}. Investigate if f(g(x)) = g(f(x)). b) The real variables y and x are related by y = 5x^2. (i) The equation y = 5x^2 can... show full transcript

Worked Solution & Example Answer:a) f(x) = 6x - 5 and g(x) = \frac{x + 5}{6} - Leaving Cert Mathematics - Question 3 - 2020

Step 1

Investigate if f(g(x)) = g(f(x))

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Answer

First, calculate f(g(x)):

  1. Substitute g(x) into f: f(g(x))=f(x+56)=6(x+56)5=x+55=x.f(g(x)) = f\left(\frac{x + 5}{6}\right) = 6\left(\frac{x + 5}{6}\right) - 5 = x + 5 - 5 = x.

Next, calculate g(f(x)):

  1. Substitute f(x) into g: g(f(x))=g(6x5)=(6x5)+56=6x6=x.g(f(x)) = g(6x - 5) = \frac{(6x - 5) + 5}{6} = \frac{6x}{6} = x.

Since both calculations yield x, we conclude:

f(g(x))=g(f(x))=x.f(g(x)) = g(f(x)) = x.

Step 2

Find the value of a and the value of b

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Answer

Rewrite the equation: y=5x2y = 5x^2 Taking logs: logby=logb(5x2)log_b y = log_b (5x^2) Using the logarithm property: logby=logb5+logb(x2)=logb5+2logbx.log_b y = log_b 5 + log_b (x^2) = log_b 5 + 2 log_b x. Thus, we have:

  • a = log_b 5
  • b = 2.

Step 3

Find the real values of y for which log_5 y = 2 + log_5 (\frac{126}{25} x - 1)

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Starting from: log5y=2+log5(12625x1).log_5 y = 2 + log_5 \left(\frac{126}{25} x - 1\right). Rewrite as: log5ylog5(12625x1)=2.log_5 y - log_5 \left(\frac{126}{25} x - 1\right) = 2. This leads to: log5(y12625x1)=2.log_5 \left(\frac{y}{\frac{126}{25} x - 1}\right) = 2. Converting to exponential form: y12625x1=52=25.\frac{y}{\frac{126}{25} x - 1} = 5^2 = 25. Rearranging gives: y=25(12625x1)=126x25.y = 25 \left(\frac{126}{25} x - 1\right) = 126x - 25. This is the function relating y and x. If a quadratic form emerges, find the zeros to determine the real values.

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