Composition of Functions Practice 3 – Visualizing Algebra

October 1, 2019

Here the solutions are, negative 23, positive 5 and negative 4x squared minus 6x plus 5. Excellent thinking, if you got all those correct. For this first one, we’ll start by finding the f of 2. We’ll take 2 and plug that in, for x in the function f. Doing so, we get an output of positive 12. So, when the input of f is 2, the output is 12. So, we just replace f of 2 with 12. Now we just need to find g of 12. We let g’s input be 12, so negative 2 times 12 is negative 24, plus 1 is negative 23. This is how we get our first value. For the second problem, we want to find g of f of 0. So, we start by finding f of 0. You let x equals 0 for the function f, and then when we solve this, or simplify it, we get negative 2. So, we know f of 0 equals negative 2. So, we just replace this value with negative 2. So, letting x equal negative 2 for g, we get an output of positive 5. This is our second solution. And finally for g of f of x, we simply take the entire function f of x, and we plug it in as the input for g of x. So, wherever we see f of x, will replace that with 2x squared plus 3x minus 2, distribute the negative 2 will have negative 4x squared minus 6x plus 4. And then we’ll have plus 1 on the end. Combing these two like terms will have our final composite function.

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