I am currently trying to find the roots of this function below.
$$ f_s(c,s,\alpha,\rho, \gamma, us) =(\alpha s^{\rho-1} ((1-\alpha)c^\rho + \alpha s^\rho)^{((1-\gamma)/\rho) - 1}) - us $$
, where us is one whole variable.
I want to find the roots of this function, given fixed values of s,alpha,rho,gamma and us. The issue is, I have typed this function into wolfram trying different fixed values, but the result is always that it yields no roots.
s = 40, alpha = 0.5, rho = 0.5, gamma = 2, us = 3
s = 20, alpha = 0.2, rho = 0.5, gamma = 2, us = 2
Is there something I'm missing or does this equation truly have no roots? I've attached some more images below that might be of help, thanks a lot!
Wolfram Alpha Giving the Origin of that function above, and us is just a new variable I made up to subtract from it.
EDIT 1 : Thank you for all your suggestions; I have tried graphing it, and for sure there are no roots. This is supposed to work with either of the fixed values I chose, so I would conclude that this function really has no roots and I was down the wrong track. If anyone else has any insights, that would be great but if not, thanks for your help everyone and I'll close this thread.
Let's rename $us$ to $u$, since it's more common to work with single-character variable names in math (and it makes it easier to read). The equation is $$\alpha s^{\rho-1}\big[(1-\alpha)c^\rho+\alpha s^\rho\big]^{\frac{1-\gamma}\rho-1}=u.$$ Now, for every quintuple $(\alpha,s,c,\rho,\gamma)$, there is (exactly) one value of $u$ for which the equation is satisfied. This means that, for at least some values of $(\alpha,s,\rho,\gamma,u)$, there is some value $c$ that satisfies the equation: pick some real $c_0$ and some parameters $(\alpha,s,\rho,\gamma)$, set $$u_0=\alpha s^{\rho-1}\big[(1-\alpha)c_0^\rho+\alpha s^\rho\big]^{\frac{1-\gamma}\rho-1},$$ and then $c=c_0$ will be a root of the equation with parameters $(\alpha,s,\rho,\gamma,u_0)$.
This may not be that satisfying, since it doesn't tell you much about when a solution might exist. Let's try to solve the system. We can first divide by $(\alpha s^{\rho-1})$ to get $$\big[(1-\alpha)c^\rho+\alpha s^\rho\big]^{\frac{1-\gamma}\rho-1}=\frac{u}{\alpha s^{\rho-1}}.$$ Now, we can take each side to the power of $\frac{1-\gamma-\rho}{\rho}$ to get $$(1-\alpha)c^\rho+\alpha s^\rho = \left(\frac{u}{\alpha s^{\rho-1}}\right)^{\frac{1-\gamma-\rho}\rho}.$$ This gives $$c^\rho=\frac1{1-\alpha}\left(\frac{u}{\alpha s^{\rho-1}}\right)^{\frac{1-\gamma-\rho}\rho}-s^\rho\frac{\alpha}{1-\alpha}.$$ So, depending on your value of $\rho$, this might be solvable (if $\rho$ is an odd integer, there is a root somewhere, while otherwise you probably want the right side to be positive). In general, though, if $$\left(\frac{u}{\alpha s^{\rho-1}}\right)^{\frac{1-\gamma-\rho}\rho}\geq \alpha s^\rho,$$ then there is a (positive real) solution for $c$.
Edit (to explain the remark on how the solvability depends on $\rho$): Say $\rho$ is some arbitrary real number. Where is $c^\rho$ defined? A priori, it's hard to say how to define exponentation exactly -- it's easy when $\rho$ is an integer and we can just multiply $c$ together a bunch of times, but harder for arbitrary $\rho$. It turns out (and Desmos, for example, will give you a graph of this) that you can define $c^\rho$ for any real $\rho$, as long as $c$ is positive (for an example of why this constraint is reasonable, think about $(-1)^{1/2}$; it turns out we can define exponentiation $x^y$ for all reals, but we lose some nice properties, like the fact that $x^y$ is a real number). These values are always positive, so there is only a solution to $c^\rho=x$ in $c$ for arbitrary $\rho$ when $x$ is positive.
If we're in the special case where $\rho$ is an integer, then we can define $c^\rho$ for all real $c$, and then try to solve for $c$. If $\rho$ is even, $c^\rho\geq 0$ for all $c$, and so we can't solve $c^\rho=x$ for negative $x$. However, if $\rho$ is odd, we can!