# State if the statement is True or False: The maximum value of $2x^3-9x^2-24x-20$ is $-7$.

State if the statement is True or False: The maximum value of $$2x^3-9x^2-24x-20$$ is $$-7$$.

Let $$f(x) = 2x^3-9x^2-24x-20$$.

If we go by the derivative test: $$f'(x) = 6x^2-18x-24 \ \ \& \ \ f'(x) = 0 \implies x=4,-1$$

At $$x=-1$$ we get $$f(x)=-7$$ and at $$x=4$$ we get $$f(x) = -132$$, so we have maximum value $$-7$$ by this method.

But this is a polynomial function, its value tends to infinity as $$x \to \infty$$.

So what can be said about the truth of the statement?

• Every local maximum $x$ of $f$ has $f'(x) = 0$, since $\frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded. – anomaly Nov 22 '18 at 2:44
• $f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins. – Phil H Nov 22 '18 at 3:03 • When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false. – Deepak Nov 22 '18 at 3:19 ## 5 Answers Hint: Let $$x=10^{20}$$. That should do it. • x=10 will also give positive value! – user8795 Nov 22 '18 at 2:49 • Well, there's your answer, @user8795. – Shaun Nov 22 '18 at 2:51 • I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is. – Shaun Nov 22 '18 at 2:55 $$f^{\prime\prime}(-1)=12(-1)-18=-30<0$$. Thus $$f$$ has a local maximum at $$-1$$ and the local maximum value is $$f(-1)=-7$$. But it does not mean $$f$$ can not assume any value greater than $$-7$$. In fact the function $$f$$ is clearly unbounded. Here $$f^{\prime}(x)>0$$ for all $$x\in (-\infty,-1)$$ and for all $$x\in (4,\infty)$$ and $$f^{\prime}(x)<0$$ for all $$x\in (-1,4)$$. Thus $$f$$ is strictly increasing in $$(-\infty,-1)\cup (4,\infty)$$ and is strictly decreasing in $$(-1,4)$$. At $$-1$$ it has a local maximum and at $$4$$ it has a local minimum. • so the conclusion is false? – user8795 Nov 22 '18 at 3:15 • Unless the word "local" is mentioned, it is certainly false. – Anupam Nov 22 '18 at 3:17 Assume the opposite: $$2x^3-9x^2-24x-20>-7$$ $$\to 2x^3-9x^2-24x-13>0$$ $$\to (2x-13)(x+1)^2>0$$ Pretty blatant from this. • Your use of implication ("$\to\$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear. – Shaun Nov 23 '18 at 3:47

For large enough $$x$$, the polynomial grows unboundedly, therefore no finite maximum exists for the function. The points then you found, are just local maximum or minimum not global

The differentiation & setting of the derivative to zero yields stationary points. A range-of-x ought to have been specified for this question, as without it it is ill-posed.

To say that the function has a maximum value of ∞ at x=∞ is a scrambling of the meaning of what's going-on, which is that the function increases without limit as x increases without limit, & therefore does not have a maximum value. If a range of x had been given, you would compare the -7 gotten as a local maximum to the value taken by the function at the upper end of the range of x (you need not consider the lower end of the range of x as the function is decreasing without limit in that direction) ... and whichever were the greater would be the answer to the question.

A fussy little point though: but it's important in many kinds of problem: the range would have to be a closed range - ie of the form $$a\leq x\leq b ,$$ often denoted $$x\in[a,b],$$ rather than what is called an open range $$a< x< b ,$$ often denoted $$x\in(a,b) ;$$ as if it were the latter that were specified, you would technically still have the problem of the non-existence of a maximum value of the function if the upper limit of the range were such that it could exceed -7 for x still less than it.