Hi,
Could you help me with a graphing question attached?
I just don't understand why when 1/x is multiplied by f(x)approaching y= infinity, when x approaches infinity on the new graph, the new graph doesn't approach y=0?
Thanks
It is
not true to assume things like \( \infty\times 0 = 0\), or that \( \frac{\infty}{\infty} = 0\).
\[ \text{For this particular graph, it is not too hard to see why.}\\ \text{The reason is because }f(x)\text{ is a cubic.} \]
More specifically, you could guess that \( f(x) = (x+3)(x-1)^2\). But the following explanation shows that for any cubic \(f(x) = ax^3+bx^2+cx+d\), as \(x\to \infty\), \(y\to \infty\) as well.
\[\text{If we take any cubic } f(x) = ax^3+bx^2+cx+d\\ \text{Then }f(x) \times \frac{1}{x} = ax^2+bx+c+\frac{d}{x}. \]
\[ \text{As }x\to \infty,\\ \text{whilst it is certainly true that }\frac{d}{x}\to 0,\\ \text{the terms }ax^2\text{ and }bx\text{ still tend to }\infty\text{ as well.}\]
\[ \text{And the sum of something tending to infinity, plus another tending to infinity,}\\ \text{plus of course the constant and the 0 term}\\ \text{should still also tend to infinity, as required.}\]
Note that this explanation can be generalised for
all polynomials with degree at least 2. If you multiplied something linear, i.e. \(f(x)=ax+b\) to \(\frac1x\), then this does not happen. But for any higher powers, this will always be the case.
Note also that if we multiplied \(f(x) = ax^3+bx^2+cx+d\) to \( \frac{1}{x^3}\), then as \(x\to \infty\), \(y \to a\). And if we multiply \(f(x)\) to \( \frac{1}{x^n}\), where \(n > 3\), then \(y\to 0\).
If you are genuinely interested in some intuition behind why
The reason more or less is because \(x^3\) grows at a much more significant rate than \(x\) grows. Therefore \(x^3\) grows much faster than \( \frac{1}{x}\) actually decays.
The growth of \(x^3\) is said to dominate the decay of \( \frac{1}{x} \). Hence as \(x\to \infty\), \(y\to \infty\) as well.
Home
Help
Search
Login
