Mapping the interval [0,1] to [-1,1] with side conditions

Question

I'm wondering if there is a differentiable function that takes $\lambda$ as parameter $$f_\lambda : [0,1] \rightarrow [-1,1], x \mapsto f(x)$$ which is monotonic growing and has the side conditions $$f_\lambda(\lambda) = 0 \wedge f_\lambda(0) = -1 \wedge f_\lambda(1)=1$$ where $\lambda \in I$ with $I \subseteq (0,1) \setminus{\frac{1}{2}}$.

I tried to find functions of the form $$ f_\lambda(x) = \sum_{i=0}^n a_ix^i$$ and $$ f_\lambda(x) = a\cdot\log(bx + c) +d $$ but could not come up with a solution for either of those.

Any ideas?

Edit:

The solution proposed by @A.J. $$ f_\lambda(x) = 2 \cdot x^{\tfrac{\ln(1/2)}{\ln(\lambda)}} - 1$$ is a solution for all $\lambda \in (0,1)$

Answer 1

One simple way would be to find any non-linear increasing function that maps $[0,1]$ to $[0,2]$, then subtract $1$; e.g. $\;2x^n-1 \;(n\neq0),\; 3^x-2,\; 2\sin(\frac{\pi}{2}x)-1,\;$ etc.

Answer 2

$f:[0,1]\rightarrow [a,b] : x\mapsto (1-x)a + xb$.

Answer 3

$$ f(x)\ :=\ \left(\frac x{1-\lambda}+\frac{1-x}\lambda\right)\cdot(x-\lambda) $$

This function is increasing in $\ [0;1]\ $only when $\ \lambda\ $ belongs to an interval neighborhood of $\ \frac 12,\ $ but not outside that neighborhood. Thus this is only a partial solution.