The graphs of the functions y=f(x)=(x-3)^2+2 and y=g(x)=1/2x+1/2 are show in the standard (x,y) coordinate plane below. Which of the following is NOT true?
F. |f(x)| = f(x)
G |g(x)| = g(x)
H. f(3) = g(3)
J. f(3.5) = g(3.5)
K. f(g(1)) = 6
The answer is G. How do I solve this problem? By the way, the f(x) function is a parabola and the g(x) function is a straight, positive line.
The answer is G. The function g(x)= 1/2x +1/2 will output a negative number if you input a negative number for x. The absolute value of g(x) will put out a positive number no matter the X value, while the original function will produce a negative number if X is less than 0
@TQconfidential what do the absolute values in F and G mean? That if any x value is put in the outcome will be positive?
@Venomyth remember absolute value just means distance from zero. If f(x) is a function, then for any given x in the domain of f, |f(x)| is simply the absolute value of f(x). Note that absolute values aren’t necessarily positive – they are non-negative. But in this case |f(x)| is always positive.
The reason why G is correct is because it is not true that |g(x)| = g(x) for all values of x, as @TQconfidential said (however, you have to feed in a number less than -1 for the statement not to hold).
@MITer94 Thanks for the explanation! Could you also help me on this one?
The solution set for the equation 2^x^2 + 1 = 1 contains:
F. 2 imaginary numbers
G. 2 positive real numbers
H. 1 negative and 1 positive real number
J. 1 negative real number only
K. 1 real number, which is 0
I know it’s F but what’s the other imaginary number if i is one of them?
@Venomyth do you mean 2^{x^2 + 1} = 1?
i is definitely a solution since it is a solution to x^2 + 1 = 0. But -i is also a solution.
@MITer94 ohhh alright for some reason I thought -i^2 would equal 1 but two negatives make a positive, thanks again