Answer:
Step-by-step explanation:
We have the two functions:
And we wish to find:
First, let’s find f(2) first. So, we will substitute 2 for x for f(x):
Hence, we can now substitute 3 for f(2):
Now, we can find g(6). Substitute 6 for t for g(t):
Therefore:
Answer:
Step-by-step explanation:
I like to use a graphing calculator for finding the zeros of higher order polynomials. The attachment shows them to be at x = -3, -1, +2.
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The zeros can also be found by trial and error, trying the choices offered by the rational root theorem: ±1, ±2, ±3, ±6. It is easiest to try ±1. Doing so shows that -1 is a root, and the residual quadratic is ...
x² +x -6
which factors as (x -2)(x +3), so telling you the remaining roots are -3 and +2.
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For any odd-degree polynomial with a positive leading coefficient, the sign of the function will match the sign of x when the magnitude of x gets large. Thus as x approaches negative infinity, so does f(x).
