Python 2.7, 34 bytes

lambda x:max(x,[18,1.4]['.'in`x`])

JavaScript (ES6), 27 31

Input taken as a string. To check if the input value has decimals, it's appended to itself: if there is no decimal point the result is still a valid number, else it's not. To recognize a valid number (including 0), I use division as in javascript 1/n is numeric and not 0 for any numeric n (eventually the value is Infinity for n==0), else it's NaN

x=>x<(y=1/(x+x)?18:1.4)?y:x

Test

f=    
x=>x<(y=1/(x+x)?18:1.4)?y:x

;[
 ['0', '18' ],['1', '18' ],['2', '18' ],['12', '18' ],['18', '18' ],['43', '43' ],['115', '115'], ['122', '122' ]
,['0.0', '1.4'],['1.0', '1.4'],['1.04', '1.4'],['1.225', '1.4'],['1.399', '1.4'],['1.4', '1.4'],['1.74', '1.74'],['2.0', '2.0'],['2.72', '2.72']
].forEach(t=>{
  var i=t[0],k=t[1],r=f(i)
  console.log(i,k,r,k==r?'OK':'KO')
})

My prev (wrong) solution:

Taking input as a number, you can use remainder operator % to check if the number is integer.

x=>x<(y=x%1?1.4:18)?y:x

or

x=>Math.max(x,x%1?1.4:18)

But this does not work as the challenge request to discriminate between, say, 2 and 2.0 and that's the same number. So it's not possibile to get the input as a number

05AB1E, 13 11 bytes

Uses CP-1252 encoding.

ÐîQ18*14T/M

Explanation

Ð             # triplicate input
 î            # round up
  Q           # check for equality
   18*        # multiply 18 by this (18 if input is int, else 0)
      14T/    # push 14 / 10
          M   # take max of stack (input and 1.4 or 18)

Try it online

Java 7, 90 bytes

String c(String i){return""+(i.contains(".")?new Float(i)<1.4?1.4:i:new Long(i)<18?18:i);}

Ungolfed & test cases:

Try it here.

class M{
  static String c(String i){
    return ""+(i.contains(".")
                ? new Float(i) < 1.4
                   ? 1.4
                   : i
                : new Long(i) < 18
                   ? 18
                   : i);
  }

  public static void main(final String[] a) {
    System.out.println(c("0"));
    System.out.println(c("1"));
    System.out.println(c("2"));
    System.out.println(c("12"));
    System.out.println(c("18"));
    System.out.println(c("43"));
    System.out.println(c("122"));

    System.out.println(c("0.0"));
    System.out.println(c("1.04"));
    System.out.println(c("1.225"));
    System.out.println(c("1.399"));
    System.out.println(c("1.4"));
    System.out.println(c("1.74"));
    System.out.println(c("2.0"));
    System.out.println(c("2.72"));
  }
}

Output:

18
18
18
18
18
43
122
1.4
1.4
1.4
1.4
1.4
1.74
2.0
2.72

PHP, 40 Bytes

modified by @user59178 Thank You

<?=max(is_int(0+$i=$argv[1])?18:1.4,$i);

PHP, 42 Bytes first Version

<?=max(strpos($i=$argv[1],".")?1.4:18,$i);

Pyth, 14 bytes

eS,@,18h.4}\.`

Try it online.

EXCEL: 26 31 Bytes

=MAX(A1;IF(MOD(A1;1)=0;18;1,4))

Formula can go anywhere except A1, the input cell.

Fixed errors, and replaced with Emigna's suggestion.

Perl, 29 27 bytes

Includes +2 for -lp

Give input on STDIN

adult.pl <<< 1.24

adult.pl:

#!/usr/bin/perl -lp
$_>($a=/\./?1.4:18)or*_=a

If you don't mind an extra newline if you really were a full adult then leaving out the l option for 26 bytes works too

GNU sed, 40 + 1 = 41 bytes

(score +1 for use of -r flag to interpreter)

s/^.$|^1[^9]$/18/
/^0|1\.[0-3]/s/.*/1.4/

Annotated:

#!/bin/sed -rf

# First, anything that's a single digit or is '1' followed by a
# digit other than '9' is replaced with '18'.
s/^.$|^1[^9]$/18/

# Now, any line beginning with '0' or containing '1.0' to '1.3' is
# replaced with '1.4'.
/^0|1\.[0-3]/s/.*/1.4/

We take advantage of the constraints on input, so don't have to test the start of string when we see '1.' - we know there's only one digit before the decimal point.

Test result:

$ ./94832.sed <<END
> 0
> 1
> 2
> 12
> 18
> 43
> 122
> 
> 0.0
> 1.04
> 1.225
> 1.399
> 1.4
> 1.74
> 2.0
> 2.72
> END
18
18
18
18
18
43
122

1.4
1.4
1.4
1.4
1.4
1.74
2.0
2.72

Brachylog, 14 bytes

#$:18ot|:1.4ot

Try it online!

Explanation

    #$             Input is an integer
      :18o         Sort the list [Input, 18]
          t        Take the last element
|              Or
    :1.4o          Sort the list [Input, 1.4]
         t         Take the last element

CJam, 18 14 bytes

Saved 4 bytes thanks to Martin Ender.

q~_`'.&1.4I?e>

Online interpreter

C#, 69 bytes

s=>s.Contains(".")?float.Parse(s)<1.4?"1.4":s:int.Parse(s)<18?"18":s;

Try it online!

Full program with test cases:

using System;

namespace YesImAnAdult
{
    class Program
    {
        static void Main(string[] args)
        {
            Func<string,string>f= s=>s.Contains(".")?float.Parse(s)<1.4?"1.4":s:int.Parse(s)<18?"18":s;

            Console.WriteLine(f("0"));  //18
            Console.WriteLine(f("1"));  //18
            Console.WriteLine(f("2"));  //18
            Console.WriteLine(f("12")); //18
            Console.WriteLine(f("18")); //18
            Console.WriteLine(f("43")); //43
            Console.WriteLine(f("122"));    //122

            Console.WriteLine(f("0.0"));    //1.4
            Console.WriteLine(f("1.04"));   //1.4
            Console.WriteLine(f("1.225"));  //1.4
            Console.WriteLine(f("1.399"));  //1.4
            Console.WriteLine(f("1.4"));    //1.4
            Console.WriteLine(f("1.74"));   //1.74
            Console.WriteLine(f("2.0"));    //2.0
            Console.WriteLine(f("2.72"));   //2.72
        }
    }
}

A pretty straightforward solution. Note that on some systems float.Parse() might return incorrect results. Pass CultureInfo.InvariantCulture as the second argument according to this answer.

Haskell, 50 bytes

x#y=show$max x$read y 
f s|elem '.'s=1.4#s|1<2=18#s

Usage example: f "1.0" -> "1.6".

Haskell's strict type requires usage of strings as input and output. Howevers, read, max and show are polymorphic and handle all numeric types.

C++, 68 bytes

int A(int a){return a<18?18:a;}float A(float h){return h<1.4?1.4:h;}

This answer is actually 2 functions with the same name, and the compiler works out for me which to call, so it acts as one function without me having to take in one input and decide which it is. Since input on the floating point is guaranteed to be of the same precision as the output, I can return it safely without having to truncate it either.

Ungolfed + tests

#include <iostream>

int A(int a)
{
   return a < 18 ? 18 : a;
}

float A(float h)
{
   return h < 1.4 ? 1.4 : h;
}

int main()
{
  std::cout << 0 << " " << A(0) << "\n";
  std::cout << 19 << " " << A(19) << "\n";
  std::cout << 1.1 << " " << A(1.1f) << "\n";
  std::cout << 2.2 << " " << A(2.2f) << "\n";
}

Actually, 16 bytes

;:.7τ9τ($'.íuIkM

Try it online!

Explanation:

;:.7τ9τ($'.íuIkM
;                 dupe input
 :.7τ             1.4 (.7*2) - note that :1.4 is the same length, but an additional delimiter would be needed to separate it from the following 1
     9τ           18 (9*2)
       ($'.íu     1-based index of "." in string representation of input, 0 if not found
             I    1.4 if input contains a "." else 18
              kM  maximum of remaining values on stack 

IBM/Lotus Notes Formula Language, 58 49 bytes

@If(@Like(@Text(a);"%.%");@If(a<1.4;1.4;a);@If(a<18;18;a))

Computed field formula where a is an editable numeric field.

EDIT

@If(@Like(@Text(a);"%.%");@Max(1.4;a);@Max(18;a))

Alternative inspired by @Mego

Dyalog APL, 14 bytes Rendered invalid by further specifications

⎕IO←0 which is default on many systems. Takes string as argument.

⍎⌈18 1.4⊃⍨'.'∘∊

⍎⌈ max of the evaluated argument and

18 1.4⊃⍨ {18,1.4} selected by

'.'∘∊ whether the argument contains a period

C#, 95 Bytes

Golfed:

string y(string p){int a;return int.TryParse(p,out a)?a>17?p:"18":double.Parse(p)<1.4?"1.4":p;}

Ungolfed:

class YesOfCourseImAnAdult
  {
    public string y(string p)
    {
      int a;
      return int.TryParse(p, out a) ? a > 17 ? p : "18"
       : double.Parse(p) < 1.4 ? "1.4" : p;
    }
  }

Test cases:

var codeGolf = new YesOfCourseImAnAdult();
Console.WriteLine(codeGolf.y("0"));
Console.WriteLine(codeGolf.y("1"));
Console.WriteLine(codeGolf.y("2"));
Console.WriteLine(codeGolf.y("12"));
Console.WriteLine(codeGolf.y("18"));
Console.WriteLine(codeGolf.y("43"));
Console.WriteLine(codeGolf.y("122"));

Console.WriteLine(codeGolf.y("0.0"));
Console.WriteLine(codeGolf.y("1.04"));
Console.WriteLine(codeGolf.y("1.225"));
Console.WriteLine(codeGolf.y("1.399"));
Console.WriteLine(codeGolf.y("1.4"));
Console.WriteLine(codeGolf.y("1.74"));
Console.WriteLine(codeGolf.y("2.0"));
Console.WriteLine(codeGolf.y("2.72"));

Output:

18
18
18
18
18
43
122

1.4
1.4
1.4
1.4
1.4
1.74
2.0
2.72

Jelly, 16 15 13 bytes

ŒṘċ”.ị1.4,18»

TryItOnline
Or see all test cases, also at TryItOnline

How?

ŒṘċ”.ị1.4,18» - Main link: n
      1.4,18 - pair literals 1.4 and 18:   [1.4,18]
     ị       - index (1-based & modular ie:^  1, 0^)
  ċ          -     count (occurrences of)
   ”.        -         string "." (present in)
ŒṘ           -         string representation of n
           » - maximum (of this and n)

C, 119 111 105 99

m;f(char*s){float atof(),l=atof(s);for(m=s;*s&*s++!=46;);puts(*s?l<1.4?"1.4":m:atoi(m)>18?m:"18");}

Tested with

main(c,v)char**v;{
    f("0");
    f("1");
    f("2");
    f("12");
    f("18");
    f("44");
    f("115");
    f("122");
    f("0.0");
    f("1.04");
    f("1.225");
    f("1.339");
    f("1.4");
    f("1.74");
    f("2.0");
    f("2.72");
}

Output

18
18
18
12
18
44
115
122
1.4
1.4
1.4
1.4
1.4
1.74
2.0
2.72

Batch, 102 bytes

@set/ps=
@if %s:.=%==%s% (if %s% lss 18 set s=18)else if %s:~0,1%%s:~2,1% lss 14 set s=1.4
@echo %s%

First determines whether the input is an integer by checking whether deleting all .s has any effect on the string. If it is then the value is readily compared against 18, otherwise the first and third characters are combined into a number which is compared against 14.

PHP, 36 35 bytes

<?=max($a=$argv[1],$a[1]&p?18:1.4);

Checks whether the character in the second position is a digit (binary and with p). If so, use 18 as the minimum, otherwise 1.4. Then take the bigger of the 2 numbers.

Run like this:

echo '<?=max($a=$argv[1],$a[1]&p?18:1.4);' | php -- 11 2>/dev/null;echo

Explanation

Here's a table of the possible results of the &"p" operation:

input  binary      &   "p"         result      chr   Truthy/falsy
"."    0010 1110   &   0111 0000   0010 0000   " "   TRUTHY
"0"    0011 0000   &   0111 0000   0011 0000   "0"   FALSY
"1"    0011 0001   &   0111 0000   0011 0000   "0"   FALSY
"2"    0011 0010   &   0111 0000   0011 0000   "0"   FALSY
"3"    0011 0011   &   0111 0000   0011 0000   "0"   FALSY
"4"    0011 0100   &   0111 0000   0011 0000   "0"   FALSY
"5"    0011 0101   &   0111 0000   0011 0000   "0"   FALSY
"6"    0011 0110   &   0111 0000   0011 0000   "0"   FALSY
"7"    0011 0111   &   0111 0000   0011 0000   "0"   FALSY
"8"    0011 1000   &   0111 0000   0011 0000   "0"   FALSY
"9"    0011 1001   &   0111 0000   0011 0000   "0"   FALSY
null   0000 0000   &   0111 0000   0000 0000   0     FALSY

When the input is 1 digit, then taking the second digit will result in null. null&"string" will also conveniently evaluate to false. So only when the second char is a . will the expression evaluate to true and 1.4 will be used as the minimum instead of 18.

PHP: 40 bytes

$i=is_int($i)?$i>17?$i:18:$i>1.4?$i:1.4;

psuedocode, (nested turnary) :

if (i is an integer) then 
  if (i is bigger than 17) then i=18 else i=i  
otherwise (its a decimal)   
  if (i is bigger than 1.4) then i=i else i=1.4 
end if 

Powershell : 58 Bytes

Bit longer than i'd expected, so many brackets required to have it run correctly.

.({({1.4},$i)[$i-gt1.4]},{({18},$i)[$i-gt18]})[$i-is[int]]

Input as $i

. #Execute the code.. required.
( #First block
    { #Value if-false of first block
        (
        {1.4} #Return Val
        , #Or
        $i #Return Input
        )[$i-gt1.4] #If input is greater than Val
    },{ #Next Block
    ({18},$i)[$i-gt18]} #Same with different val
)[$i-is[int]] #Decides based on if val is default int

Test Cases:

@(0,1,2,12,18,43,115,122,0.0,1.04,1.225,1.399,1.4,1.74,2.0,2.72) | % {
$i = $_
Write-Host $i`t`t -NoNewline
.({({1.4},$i)[$i-gt1.4]},{({18},$i)[$i-gt18]})[$i-is[int]]
}

returns :

0       18
1       18
2       18
12      18
18      18
43      43
115     115
122     122
0       1.4
1.04    1.4
1.225   1.4
1.399   1.4
1.4     1.4
1.74    1.74
2(.0)   2      (this is a value of type 'double' - the default formatting of powershell doesn't add decimals, but the returned value is of the correct type)
2.72    2.72

Proof of 2 == 2.0

$i = 2.0
$q = .({({1.4},$i)[$i-gt1.4]},{({18},$i)[$i-gt18]})[$i-is[int]]
$q.GetType().Name
Double

Python 2, 49 bytes

lambda a:(a,1.4)[a<1.4]if"."in`a`else(a,18)[a<18]

Java, 95 bytes

void p(double d){if(d%1>0)System.out.print(d<1.4?1.4:d);else System.out.print(d<18?18:(int)d);}

5 worse than Kevin's answer, but it has the print statement inside the function so maybe I get some pity points for that.

If this is allowed in Java, you could also use

d->{if(d%1>0)System.out.print(d<1.4?1.4:d);else System.out.print(d<18?18:d);};

Which could be used like

Consumer<Double> func = d->{if(d%1>0)System.out.print(d<1.4?1.4:d);else System.out.print(d<18?18:d);};

C, 50 bytes:

#define A(x)(x/2+(x+1)/2-x?x<1.4?1.4:x:x<18?18:x)

The byte count includes the newline at the end of the macro definition.

Test:

#define A(x)(x/2+(x+1)/2-x?x<1.4?1.4:x:x<18?18:x)
#include <assert.h>
int main() {
  assert(A(0) == 18);
  assert(A(1) == 18);
  assert(A(2) == 18);
  assert(A(12) == 18);
  assert(A(18) == 18);
  assert(A(43) == 43);
  assert(A(115) == 115);
  assert(A(122) == 122);
  assert(A(0.0) == 1.4);
  assert(A(1.04) == 1.4);
  assert(A(1.225) == 1.4);
  assert(A(1.399) == 1.4);
  assert(A(1.4) == 1.4);
  assert(A(1.74) == 1.74);
  assert(A(2.0) == 2.0);
  assert(A(2.72) == 2.72);
}

C#, 58 bytes

x=>x is int?(int)x>17?x:18:(float)x<1.4?"1.4":$"{x:.0##}";

No crazy string parsing needed for C#. Input is expected to be an int or float (unfortunately C# can't cast a double to float if the double is in an object). Output will be either int or string in an object.

(almost missed the at least 1 decimal requirement, added that now)

Ungolfed:

/*Func<object, object> Lambda = */ x =>
    x is int // if parameter is an int
        ? (int)x > 17 // check if x is at least 18
            ? x // at least 18 so return x
            : 18 // less than 18 so return 18
        : (float)x < 1.4 // x is float, check if at least 1.4
            ? "1.4" // less than 1.4 so return 1.4
            : $"{x:.0##"} // at least 1.4 so return x and ensure at least 1 decimal place
;

Alternate implementation which is also 58 bytes.

x=>x is int?(int)x>17?x:18:$"{((float)x<1.4?1.4:x):.0##}";

C#, 71 bytes

object A(object i){return i is int?(int)i>18?i:18:(double)i>1.4?i:1.4;}

try it here

Emacs Lisp, 37 bytes

(lambda(x)(max(if(floatp x)1.4 18)x))

Guesses from the "datatype" whether the integer or float version should be used. (floatp returns t for 1.0, but not for 1.) The parameter is a number as integer or float, i.e. it should satisfy numberp.