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Let $f$ be a twice differentiable function defined in $[-3,3]$ such that $f(0)=-4, f’(3)=0, f’(-3)=12$, and $f’’(x)\geq-2, \forall x \in [-3,3]$. If $g(x)=\int_0^x {f(t)dt}$ then find the maximum value of g(x).

The thing is that I’m trying to get some info out of this about the function $f$, by basically having a rough graph, but the criterion $f’’(x)\geq-2$ doesn’t really give me anything because $f$ could be either convex-up(like $x^2$) or convex-down(like $-x^2$) or even have a point of inflection. It just can’t be too-much-concave-down, if you know what I mean. The function starts off as increasing, but IDK what happens after that. Also if it were just $\int_0^x {f’(t)dt}$ I might have been more confident, but $\int_0^x {f(t)dt}$ just throws me off.

Feng
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insipidintegrator
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    Hint: $,f',$ is a function that drops by $,12,$ between $,-3,$ and $,3,$ but its derivative is $,\ge -2,$ at each point. – dxiv Jun 04 '22 at 07:00
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    Oooh nice! Lemme see…so f”(x)=-2 right?because if it goes above -2 it must also fall below -2 at some point right? How to formalise this? – insipidintegrator Jun 04 '22 at 07:02
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    Formalize this using inequalities on the quantity $\int_{-3}^{3} f''(x)dx$. Use a familiar result on integral inequalities to show that $f''$ must be equal to $-2$ everywhere. (If you still find it unfamiliar, then look at $\int_{-3}^{3} (f''(x)+2) dx$ instead) [When is the integral of a non-negative function equal to zero?] – Sarvesh Ravichandran Iyer Jun 04 '22 at 07:03
  • Which result! I’m not familiar with many – insipidintegrator Jun 04 '22 at 07:04
  • Ah,@SarveshRavichandranIyer I saw your edit. Thanks – insipidintegrator Jun 04 '22 at 07:10
  • @insipidintegrator draw a straight line from $(-3,12)$ to $(3,0)$. The slope of this is $-2$. Now draw a curve between these two points that has a slope greater than that of the line everywhere in that interval. If you are unable to then the slope must be equal to slope of line. – sku Jun 04 '22 at 07:17
  • @SarveshRavichandranIyer can you please post your comment as an answer? I’ll accept it and this question can be done away with. – insipidintegrator Jun 04 '22 at 08:00
  • @sku Yes this was what I wanted to formalise. It has been done, and I have requested @ SarveshRavichandranIyer to post it as an answer. – insipidintegrator Jun 04 '22 at 08:02
  • @insipidintegrator Sure, I'll post it as a complete answer. – Sarvesh Ravichandran Iyer Jun 04 '22 at 11:40

1 Answers1

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Consider the (well defined, because $f''$ is integrable by the fundamental theorem of calculus) quantity $$ \int_{-3}^{3} (f''(x)+2) dx=S $$

On the one hand :

  • Note that $f''(x) + 2 \geq 0$ on the interval $[-3,3]$

  • An anti-derivative of $f''(x)+2$ is $f'(x) + 2x + C$ using the fundamental theorem of calculus, therefore $$ S = [f'(x)+2x+C]^{3}_{-3} = f'(3) - f'(-3) + 12 = 0 $$ by the assumptions on $f'(3),f'(-3)$.

All in all, it follows that $$ \int_{-3}^{3} h(x) = 0 $$ where $h$ is a non-negative function on $[-3,3]$. By a well-known result, it follows that $h= 0$ a.e. (almost everywhere), or that $f'' = -2$ a.e.

Next, integrating this equality from $-3$ to $x$ for $x \in [-3,3]$, we get $$ \int_{-3}^x f''(t)dt = \int_{-3}^x (-2)dt \implies f'(x) - f'(-3) = -6-2x \implies f'(x) = 6-2x $$

(There's no a.e. needed because $f'$ is necessarily continuous being the integral of another function) Integrating that equality from $0$ to $y$ for $y \in [-3,3]$ gives $$ \int_{0}^y f'(x)dx = \int_{0}^y (6-2x)dx \implies f(y) - f(0) = 6y-y^2 \implies f(y) = 6y-y^2-4 $$ Note that $g'(y) = f(y)$, therefore at a point of maximum we must have that $f(y) = 0$ (using a common result that the derivative must be zero at any extremum). Now, $f(y) = 0$ is a quadratic equation having the roots $3 \pm \sqrt{5}$. Only $3-\sqrt{5}$ lies in the interval, and since $g''(y) = f'(y)$, we note that $$ f'(3 - \sqrt{5}) = 6-2(3-\sqrt{5}) = 2\sqrt{5} > 0 $$ Therefore, ironically enough, by the double derivative test, this point is a point of minimum, and not a maximum. This is confirmed when one sees that $g(3-\sqrt{5}) < 0$ by computation.

It follows that the extreme values must therefore be taken on the ends of the interval i.e. in $\{-3,3\}$ (because the derivative would be zero at an interior point). A simple examination will tell you that $g(-3)> g(-3)$. It follows that $g(-3) = 48$ is the maximum value of $g(x)$ on $[-3,3]$.

There is a shortcut that one can use to avoid the double derivative test. Note that $g''(x) = f'(x)$ is actually non-negative on $[-3,3]$. Such functions, whose second derivative is non-negative on an interval, are called convex functions. Now, as it turns out, there is a result that makes one-dimensional analysis with convex functions very appealing : convex functions on intervals always take their maximum values on the endpoints of the interval (unless they're constant, and therefore take their maximum everywhere).

Once you recognize this, you needn't even go to the derivative test : just get $g$, and find $g(3),g(-3)$. The bigger of the two is the maximum, and you're done.

Sarvesh Ravichandran Iyer
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