Array indices must be positive integers or logical values.

x = 0:0.1:4;

f = x;

g = -2*log10(0.01+10e-4*x);

plot(x,f,x,g);

func = 2*log10(0.01+10e-4*x) + x;

res = bisect(func, 0, 4, 10e-8)

function xc = bisect(f,a,b,tol)

if sign(f(a))*sign(f(b)) >= 0

error('f(a)f(b)<0 not satisfied!') % ceases execution

end

fa = f(a);

fb = f(b);

while (b-a)/2 > tol

c = (a+b)/2;

fc = f(c);

if fc == 0 % c is a solution, done

break

end

if sign(fc)*sign(fa) < 0 % a and c make the new interval

b = c; fb = fc;

else % c and b make the new interval

a = c; fa = fc;

end

end

xc = (a+b)/2; % new midpoint is best estimate

end

Answer 1

[VBBV]

x = 0:0.1:4;
f = x;
g = -2*log10(0.01+10e-4*x);
plot(x,f,x,g);

func = @(x) 2*log10(0.01+10e-4*x) + x;
res = bisect(func, 0, 4, 10e-8)

res = 3.7250

function xc = bisect(func,a,b,tol)
if sign(func(a))*sign(func(b)) >= 0
    error('f(a)f(b)<0 not satisfied!') % ceases execution
end
fa = func(a);
fb = func(b);
while (b-a)/2 > tol
    c = (a+b)/2;
    fc = func(c);
    if fc == 0 % c is a solution, done
        break
    end
    if sign(fc)*sign(fa) < 0 % a and c make the new interval
        b = c; fb = fc;
    else % c and b make the new interval
        a = c; fa = fc;
    end
end
xc = (a+b)/2; % new midpoint is best estimate
end

[VBBV]

The below line is probably the anonymous function which you wanr to evaluate using bisect function

func = @(x) 2*log10(0.01+10e-4*x) + x;