Function with 2 unknowns and one needs to be solved

Question

I have a homework question stating:

Find the set of values of $k$ for which $f(x)=3x^2-5x-k>1$ for all $x\in\mathbb R$.

This question has really confused me because I looked at the answer and the value of $k$ is meant to be $k<-\frac{37}{12}$ and I'm afraid I don't know how to get there. May I please have some help?

Answer 1

Rearranging gives $3x^2-5x-k-1>0$, i.e. the resulting quadratic in $x$ has no real roots. This means that the discriminant $b^2-4ac$ must be negative: $$(-5)(-5)-4\cdot3(-k-1)<0$$ $$25+12(k+1)<0$$ $$k+1<-\frac{25}{12}$$ $$k<-\frac{37}{12}$$ as expected.

Answer 2

Hint

Consider the function $$f(x)=3x^2-5x-k-1$$ Its derivative $f'(x)=6x-5$ is zero when $x=\frac 56$ and you want that $f\left(\frac{5}{6}\right) >0$.

Then, ....

Answer 3

Hint: Completing the square we get $$\left(x-\frac{5}{6}\right)^2-\left(\frac{25}{36}+\frac{k}{3}+\frac{1}{3}\right)>0$$