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

Question

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?

Answer 1

Hint: Let $x=10^{20}$. That should do it.

Answer 2

$f^{\prime\prime}(-1)=12(-1)-18=-30<0$. 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.

Answer 3

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.

Answer 4

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.

Answer 5

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