(Original post by jessyjellytot14)
"Prove, by counter-example, that the statement
“ln (3x 2 + 5x + 3) ≥ 0 for all real values of x” is false."
Hmmm I'm not quite sure how to go about this so i'll post my working below & could you check it please?
Also can some explain what a real number is? I don't do further maths but I'm assuming it's just a whole integer.. So for example, -2/3 isn't a real number...?
Another way of doing this is to differentiate the expression and set the first derivative to equal zero in order to find the value of x that minimises y. You can then see if this makes f(x) => 0.
Let y = ln(u) where u = (3x^2 + 5x + 3). du/dx = 6x + 5, dy/du = 1/(3x^2 + 5x + 3).
so dy/dx = (6x + 5) / (3x^2 + 5x + 3). dy/dx = 0 when 6x + 5 = 0, so 6x = -5, and x = -5/6.
Sub in x = - 5/6 and you have your proof by counter-example.
But did you bother to watch?