The graph of the function $f(x) = 3^x$, where $x \in \mathbb{R}$, cuts the y-axis at (0, 1) as shown in the diagram below - Leaving Cert Mathematics - Question 2 - 2019

To verify that $f(1.9) < g(1.9)$, we compute both functions:

1. Calculate $f(1.9)$: $f(1.9) = 3^{1.9} \approx 11.115$

2. Calculate $g(1.9)$: $g(1.9) = 4(1.9) + 1 = 7.6$

Now, comparing the two values: $f(1.9) \approx 11.115 < 7.6$

This confirms that $f(1.9) < g(1.9)$.

To prove $3^n \geq 4n + 1$ for $n \geq 2$ by induction:

1. Base case: For $n = 2$: $f(2) = 3^2 = 9 \geq g(2) = 4(2) + 1 = 9$ Thus, the base case holds.

2. Inductive step: Assume the statement holds for some $k \geq 2$, that is, assume $3^k \geq 4k + 1$.

3. We need to prove that it holds for $k + 1$: $P(k + 1): 3^{k+1} \geq 4(k + 1) + 1$ Expanding gives us: $3 \cdot 3^k \geq 4k + 4 + 1$ By the inductive hypothesis, we substitute: $3(4k + 1) \geq 4k + 5$ Simplifying: $12k + 3 \geq 4k + 5$ This rearranges to: $8k \geq 2$ Since this is true for $k \geq 2$, it concludes our proof.

Thus, by mathematical induction, $3^n \geq 4n + 1$ for all $n \geq 2$.