Investigating the real roots of a cubic equation

[brotherbobby]

Homework Statement

Show that has only one real root unless $-5\le k\le\dfrac{7}{4}$.

Relevant Equations

(1) An th order polynomial will have at least one real root if .
(2) Rolle's theorem : If is a polynomial equation and , then there exists at least one value in the interval such that .
(3) From Rolle's theorem (2) above, it follows that if our polynomial equation has three real roots, then between any consecutive pair of them there must be a real value of for which .
(4) The condition (3) above is necessary, but not sufficient. Sufficience is however guaranteed if it can be shown that for those two values of , say some , the values of the original function must have opposite signs. Thus .

Statement of the problem : Let me copy and paste the problem as it appears in the text. [Riley, K.F., Hobson,M.P., Bence,S.J. (2006) Mathematical Methods for Physics and Engineering. (3E)].

Attempt : One real root is guaranteed, as $n =\;\text{odd}$. Differentiating $f'(x) = 12x^2+6x-6=6(2x-1)(x+1)$. Thus $f'(-1) = f'(1/2) = 0$. Hence, the derivative $f'(x)$ itself has two real roots, which indicates that $f(x)$ may have three real roots.
To test whether it does, we use condition (4) in the Relevant Equations section, given above. Namely, evaluate the value of the function $f(x)$ at those two values where its derivative vanishes to see whether they are of opposite signs, i.e. whether $f(\beta_1)f(\beta_2)<0$, where $\beta_1 = -1,\;\beta_2=1/2$.
$\small{f(-1) = 5-k\;\text{and}\; f(1/2) = -7/4-k\Rightarrow (k-5)(k+7/4)<0\Rightarrow\boxed{\color{red}{-7/4\le k\le 5}}}$.

Text answer : This is different from the answer in the text, where $\boxed{\color{blue}{-5 \le k \le 7/4}}$.
The text answer is correct, as I could verify at $\verb|desmos.com|$.

Request : A hint as to where am I going wrong in my working.

Many thanks.

 

[pasmith]

Your working appears to be correct, and I obtained the same result by considering the discriminant of $f$.

What did you do to verify the text answer? According to the text, for $k = -3$, $f$ should have more than one real root, but according to Wolfram Alpha there is only one real root and a complex conjugate pair.

I must therefore conclude that the text answer is incorrect. It happens.

EDIT: It is a bit strange to have a minus sign before the $k$ in the problem statement, so maybe that was meant to be a plus sign, which would make the answer in the text correct.