I was wondering if what I did is right and is considered to be a valid approach when solving similar questions?
Here is the question
Here is what I did
I noticed that it is a question involving the triangle inequality, so I proceeded as follows: $$|\frac{x^3+x^2-1}{x-6}| \leq |\frac{x^3}{x-6}| + |\frac{x^2}{x-6}| + |\frac{-1}{x-6}|$$ $$\leq |\frac{2^3}{2-6}| + |\frac{2^2}{2-6}| + |\frac{-1}{2-6}|$$ $$=3.25$$
Also, if the expression was $|\frac{1}{x^3+x^2-1}|$ instead of $|\frac{x^3+x^2-1}{x-6}|$, would we need to follow the same procedure but instead of plugging in $x = 2$, we would plug in $x = 1 $ since $\frac{1}{2}<\frac{1}{x}<1$?
Thanks!
All your steps are correct. But you can also notice that the numerator is monotonically increasing for $1\le x \le 2$, and the absolute value of the denominator is monotonically decreasing for $1\le x \le 2$, so
$$\left|\frac{x^3+x^2-1}{x-6}\right|\le\frac{|2^3+2^2-1|}{|2-6|}=\frac{11}{4}.$$
If you have $\left|\frac{1}{x^3+x^2-1}\right|$, you can use again the fact that $x^3+x^2-1$ is monotonically increasing for $1\le x\le 2$. This means that $$1\le x^3+x^2-1\le 11$$ and the same is true for $|x^3+x^2-1|$. Taking inverses reverses the inequalities, so $$1\ge \frac{1}{|x^3+x^2-1|}\ge \frac{1}{11}.$$
Your upper bound is $1$ in this case.