# ?

Average Error: 0.0 → 0.0
Time: 2.8s
Precision: binary64
Cost: 6592

# ?

$-1000 \leq x \land x \leq 1000$
$e^{x} \cdot x$
$x \cdot e^{x}$
(FPCore (x) :precision binary64 (* (exp x) x))
(FPCore (x) :precision binary64 (* x (exp x)))
double code(double x) {
return exp(x) * x;
}

double code(double x) {
return x * exp(x);
}

real(8) function code(x)
real(8), intent (in) :: x
code = exp(x) * x
end function

real(8) function code(x)
real(8), intent (in) :: x
code = x * exp(x)
end function

public static double code(double x) {
return Math.exp(x) * x;
}

public static double code(double x) {
return x * Math.exp(x);
}

def code(x):
return math.exp(x) * x

def code(x):
return x * math.exp(x)

function code(x)
return Float64(exp(x) * x)
end

function code(x)
return Float64(x * exp(x))
end

function tmp = code(x)
tmp = exp(x) * x;
end

function tmp = code(x)
tmp = x * exp(x);
end

code[x_] := N[(N[Exp[x], $MachinePrecision] * x),$MachinePrecision]

code[x_] := N[(x * N[Exp[x], $MachinePrecision]),$MachinePrecision]

e^{x} \cdot x

x \cdot e^{x}


# Try it out?

Your Program's Arguments

Results

 In Out
Enter valid numbers for all inputs

# Derivation?

1. Initial program 0.0

$e^{x} \cdot x$
2. Final simplification0.0

$\leadsto x \cdot e^{x}$

# Alternatives

Alternative 1
Error1.1
Cost320
$x + x \cdot x$
Alternative 2
Error1.9
Cost64
$x$

# Reproduce?

herbie shell --seed 1
(FPCore (x)
:name "exp(x)*x"
:precision binary64
:pre (and (<= -1000.0 x) (<= x 1000.0))
(* (exp x) x))