Roots of a function

Question

Why does the function: $f(y)=5y+4+8\cdot2^{3y} $ have only one root? What is the example of a similar function that has more than one root?

Answer 1

Note that

$$f(y)=5y+4+8\cdot2^{3y}$$

is surjective since it is continuos and

$$\lim_{x\to \pm \infty} f(y)=\pm \infty$$

and injective since it is strictly increasing and thus more in general

$$f(y)=k$$

has an unique solution $\forall k\in \mathbb{R}$.

As an example of a similar function with more than one root let consider

$$h(y)=-100y+4+8\cdot2^{3y}$$

$$g(y)=5\sin{y}+4+8\cdot2^{3y}$$

Answer 2

you can solve this equation with the help of the LambertW function it is $$x=-1/15\,{\frac {12\,\ln \left( 2 \right) +5\,{\it LambertW} \left( 3/5 \,\ln \left( 2 \right) {2}^{3/5} \right) }{\ln \left( 2 \right) }}=-1$$

Answer 3

Your function $f$ is the sum of strictly increasing functions and therefore it must be strictly increasing. So, it can't be $0$ twice.

But $f(-1)=0$, and therefore it has one root.

Answer 4

If you want more than one root, then assuming that the exponential term is positive, the coefficient of $y$ (which is $5$ in your function) must be negative. This will also mean that it's possible to have no roots, so you will have to have a constant term ($8$ in your function) can't be too large.

$g(y) = -5y-8+8\cdot 8^y$ is one such function.

Answer 5

First of all, this function ( $f(y)=5y+4+8\cdot2^{3y}$ ) is an ever increasing function. It doesn't go down for any interval .

For a function $f(x)$ to have more than one root, it must pass through x-axis once (a root) then go up (or down) then goes down (or up) and cut x-axis again (another root) and so on.

For a function which always increases (or decreases) can only cross x-axis once, not more than that. So, it can have only one root.

Answer 6

Nit-picking: this function has a single zero- only equations have "roots". The zeros of the function f(x) are the roots of the equation f(x)= 0.

Now,

1) f is the sum of continuous functions so it is continuous.

2) f(1)= 5+ 4+ 8(8)= 73> 0 and f(-2)= -10+ 4+ 1= -5< 0. Therefore there exist at least one x such that f(x)= 0.

3) $f'(x)= 5+ 24 ln(2)2^{3x}$ is positive for all x. Therefore f is an increasing. There is only one value of x such that f(x)= 0.

Answer 7

Let me use $x$ for the independent variable.

$y=f(x)=5x+4+8\cdot2^{3x}=5x+4+8^{x+1} = 5(x+1) + 8^{x+1} -1 $ has a monotonic behaviour going through zero at $x=-1,$ it is ever (strictly) increasing. So if at all, it can have a single crossing on x-axis with only a single root.