Optimize a Solver function - r

Question

I'm trying to set up a "Solver" function to optimize the value of "gfc" to zero varying (and finding) the variable "fc" on equation below. The parameters are given.
f0 = 6
f1 = 1
k = 2
ft = 0.3
gfc = ft-((f0-fc)/k)+((f1/k)*ln((fc-f1)/(f0-f1)))
Solving this function on Excel, I found the value of fc=5.504.

Answer 1

You can use uniroot to find where a function equals zero:
f0 = 6
f1 = 1
k = 2
ft = 0.3
gfc = function(fc) {
ft - ((f0 - fc) / k) + ((f1 / k) * log((fc - f1) / (f0 - f1)))
}
uniroot(gfc, interval = c(f0, f1))
#> $root
#> [1] 5.504386
#>
#> $f.root
#> [1] 6.72753e-09
#>
#> $iter
#> [1] 5
#>
#> $init.it
#> [1] NA
#>
#> $estim.prec
#> [1] 6.103516e-05

Answer 2

I assume that what you mean is that you want to solve for the value of fc for which gfc equals zero. We assume fc lies between f0 and f1. In that case using the constants in the question we have the following base R solutions. (Additionally packages with such functionality include nleqslv and rootSolve.)
1) optimize we can minimize gfc^2:
gfc <- function(fc) ft-((f0-fc)/k)+((f1/k)*log((fc-f1)/(f0-f1)))
optimize(function(x) gfc(x)^2, c(f0, f1))
giving:
$minimum
[1] 5.504383
$objective
[1] 4.777981e-12
2) uniroot or we can do it directly using uniroot:
u <- uniroot(gfc, c(f0, f1))
giving:
> u
$root
[1] 5.504386
$f.root
[1] 6.72753e-09
$iter
[1] 5
$init.it
[1] NA
$estim.prec
[1] 6.103516e-05
3) We can also solve this directly without any function like optimize or uniroot by rewriting
gfc(fc) = 0
as this where we have moved the first term of gfc to the LHS and then isolated fc in that term putting everything else on the RHS.
fc = f0 - k*(ft + ((f1/k)*log((fc-f1)/(f0-f1))))
Writing this as:
fc = f(fc)
we just iterate f.
f <- function(fc) f0 - k*(ft + ((f1/k)*log((fc-f1)/(f0-f1))))
fc <- (f0 + f1)/2 # starting value
for(i in 1:10) fc <- f(fc)
fc
## [1] 5.504386
4) brute force Another approach is to evaluate gfc at many points and just pick the one for which gfc^2 is least. The finer you subdivide the interval the more accurate the answer.
s <- seq(f0, f1, length = 100000)
g <- gfc(s)
s[which.min(g^2)]
## [1] 5.504395
Graphics
We can show the solution:
curve(gfc, f0, f1)
abline(h = 0, v = u$root, lty = 2)
axis(1, u$root, round(u$root, 3))