Answer to Question #184139 in Real Analysis for Jonny

Question

(i) Prove by mathemarical induction on n that

3n ≥ 2n2 + 1 for all n ∈ N

(ii) Given the function g : R → R defined by

􏰂 x−1

g (x) = 2x+4 1

2

if x̸=−2 if x=−2

Find whether or not f is injective and surjective.

Find the inverse of f, if it exists.

Answer 1

(i) $3n\ge 2n^2+1$

at $n=1, 3\ge 2+1=3$

P(1) is true

$\text{At } n=k, P(k): 3k\ge 2k^2+1~~~~~-(1)$

at n=k+1,

$p(k+1)$ :$3(k+1)\ge 2(k+1)^2+1$

$3k+3\ge 2k^2+2+4k+1$

$3k\ge 2k^2+4k$

As from eqn.(1)-$3k\ge 2k^2+1 \text{ and }3k\ge 2k^2+4k$

$\Rightarrow$ The given statement $p(n):3k\ge 2n^2+1$ is true $\forall n\in N$ .

(ii)

$g(x)=2x+4$

for Every value of x, There are different value g(x) So g(x) is injective.

Since the image im(X) of f equals the codomain function g(x), So g(x) is surjective.

Hence g(x) is injective and surjective.

Inverse:

Let $g(x)=y\Rightarrow 2x+4=y\Rightarrow x=\dfrac{y-4}{2}$

Hence The inverse is- $g^{-1}(x)=\dfrac{x-4}{2}$