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?

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Feb 21 '19 at 09:51
Yes, I do know how to complete the square but all it gets me is k< 3(x-2.5)^2 - 7.5, when I try to rearrange for k — Mason Kennedy, Feb 21 '19 at 09:56

score 2 · Answer 1 · answered Feb 21 '19 at 09:55

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, ....

answered Feb 21 '19 at 09:55

Claude Leibovici

260,315

score 2 · Accepted Answer · answered Feb 21 '19 at 09:57

2

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.

answered Feb 21 '19 at 09:57

Parcly Taxel

103,344

This helped greatly, thanks! It really annoys me that I couldn't think of that, nor would I have ever probably thought of that so thank you! – Mason Kennedy Feb 21 '19 at 10:04

score 0 · Answer 3 · answered Feb 21 '19 at 09:57

0

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

answered Feb 21 '19 at 09:57

Dr. Sonnhard Graubner

95,283

Function with 2 unknowns and one needs to be solved

3 Answers3