Prove range for derivative given $3$ definite integrals

Solution 1:

Since $f$ is continuously differentiable, by intermediate value theorem, it suffices to show that $\exists c, d \in [0, 1]$ s.t. $f'(c)=-24, f'(d)=60$.

Let $g_t(x)=tx-f(x)$, so that $g_t'(x)=t-f'(x)$. Note: Since $f$ is continuously differentiable, so is $g_t$, so $g_t'$ is continuous. We have:

\begin{align} \int_{0}^{1}{g_t(x) dx}=\int_{0}^{1}{tx dx}-\int_{0}^{1}{f(x) dx}=\frac{t}{2}-1 \\ \int_{0}^{1}{xg_t(x) dx}=\int_{0}^{1}{tx^2 dx}-\int_{0}^{1}{xf(x) dx}=\frac{t}{3}-2 \\ \int_{0}^{1}{x^2g_t(x) dx}=\int_{0}^{1}{tx^3 dx}-\int_{0}^{1}{x^2f(x) dx}=\frac{t}{4}-3 \end{align}

Using integration by parts,

\begin{align} \int_{0}^{1}{xg_t'(x) dx}=g_t(1)-\int_{0}^{1}{g_t(x) dx}=g_t(1)+(1-\frac{t}{2}) \\ \int_{0}^{1}{x^2g_t'(x) dx}=g_t(1)-\int_{0}^{1}{2xg_t(x) dx}=g_t(1)+(4-\frac{2t}{3}) \\ \int_{0}^{1}{x^3g_t'(x) dx}=g_t(1)-\int_{0}^{1}{3x^2g_t(x) dx}=g_t(1)+(9-\frac{3t}{4}) \end{align}

When $t=60$,

\begin{align} &\int_{0}^{1}{x^2(1-x)g_{60}'(x) dx} \\ &=\int_{0}^{1}{x^2g_{60}'(x) dx}-\int_{0}^{1}{x^3g_{60}'(x) dx} \\ &=(g_{60}(1)+(4-\frac{2(60)}{3}))-(g_{60}(1)+(9-\frac{3(60)}{4})) \\ &=0 \end{align}

Clearly if $g_{60}'(x)>0 \, \forall x \in [0, 1]$, then $\int_{0}^{1}{x^2(1-x)g_{60}'(x) dx}>0$, a contradiction. If $g_{60}'(x)<0 \, \forall x \in [0, 1]$, then $\int_{0}^{1}{x^2(1-x)g_{60}'(x) dx}<0$, a contradiction.

Thus $g_{60}'(a) \leq 0 \leq g_{60}'(b)$ for some $a, b \in [0, 1]$ (not necessarily distinct). By intermediate value theorem, (since $g_{60}'(x)$ is continuous) $g_{60}'(d)=0$ for some $a \leq d \leq b$, so $d \in [0, 1]$ and $f'(d)=60-g_{60}'(d)=60$.

When $t=-24$, then we want to show that $g_{-24}'(c)=0$ for some $c \in [0, 1]$. Assume on the contrary that no such $c$ exists. By intermediate value theorem (since $g_{-24}'(x)$ is continuous), this is only possible if $g_{-24}'(x)$ is either always positive over $[0, 1]$ or always negative over $[0, 1]$. In any case, for $n=1, 2, 3$:

$$\int_{0}^{1}{|x^ng_{-24}'(x)| dx}=\left|\int_{0}^{1}{x^ng_{-24}'(x) dx}\right|$$

Also $\int_{0}^{1}{x^ng_{-24}'(x) dx}$ has the same sign for $n=1, 2, 3$.

Now by Cauchy Schwarz inequality,

\begin{align} & (g_{-24}(1)+(1-\frac{-24}{2}))(g_{-24}(1)+(9-\frac{3(-24)}{4})) \\ & =\int_{0}^{1}{xg_{-24}'(x) dx}\int_{0}^{1}{x^3g_{-24}'(x) dx} \\ & =\left|\int_{0}^{1}{xg_{-24}'(x) dx}\right|\left|\int_{0}^{1}{x^3g_{-24}'(x) dx}\right| \\ & =\int_{0}^{1}{|xg_{-24}'(x)| dx}\int_{0}^{1}{|x^3g_{-24}'(x)| dx} \\ & =\int_{0}^{1}{\left(\sqrt{|xg_{-24}'(x)|}\right)^2 dx}\int_{0}^{1}{\left(\sqrt{|x^3g_{-24}'(x)|}\right)^2 dx} \\ & \geq \left(\int_{0}^{1}{|x^2g_{-24}'(x)| dx}\right)^2 \\ & =\left(\left|\int_{0}^{1}{x^2g_{-24}'(x) dx}\right|\right)^2 \\ & =(g_{-24}(1)+(4-\frac{2(-24)}{3}))^2 \end{align}

Thus $(g_{-24}(1)+13)(g_{-24}(1)+27) \geq (g_{-24}(1)+20)^2$, giving $351=13(27) \geq 20^2=400$, a contradiction.

Therefore $g_{-24}'(c)=0$ for some $c \in [0, 1]$, and we immediately have $f'(c)=-24-g_{-24}'(c)=-24$.