Daily math problems for July 31, 2026 — Calculus, Logaritmi, Sequences & more

$f^{\mathrm{\prime }}(x)=3x^2-3$, $f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$. $f^{\mathrm{\prime }\mathrm{\prime }}$ changes sign at $x=0$, so the inflection is at $x=0$.

Avem $18=2\cdot 3^2$, deci ${\log }_{10}18={\log }_{10}2+{\log }_{10}{3}^2={\log }_{10}2+2{\log }_{10}3=a+2b$.

$q={\displaystyle \frac{b_2}{b_1}}={\displaystyle \frac{12}{4}}=3$.

Base along the $x$-axis has length $3$; the third vertex sits at height $4$. Area $=\frac{1}{2}\cdot 3\cdot 4=6$.

$A^2=\left(\begin{array}{cc}0\cdot 0+1\cdot 0& 0\cdot 1+1\cdot 0\\ 0\cdot 0+0\cdot 0& 0\cdot 1+0\cdot 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$. ($A$ is nilpotent.)

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{(1\mathrm{/}x)\cdot x-\ln x\cdot 1}{x^2}}={\displaystyle \frac{1-\ln x}{x^2}}$. At $x=1$: ${\displaystyle \frac{1-0}{1}}=1$.

$\bullet$ Find vertex $A=AB\cap AC$ by solving $x-2y=-3$ and $2x-y=3$:

$$A=(3,3)$$

$\bullet$ Find vertex $B=AB\cap BC$ by solving $x-2y=-3$ and $3x+2y=-1$:

$$B=(-1,1)$$

$\bullet$ Find vertex $C=AC\cap BC$ by solving $2x-y=3$ and $3x+2y=-1$:

$$C=(\frac{5}{7},-\frac{11}{7})$$

$\bullet$ Apply the shoelace-style area formula:

$$\mathcal{A}=\frac{1}{2}\mid x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\mid$$

$\bullet$ Plugging the coordinates gives $\frac{1}{2}\cdot \frac{96}{7}$:

$$\mathcal{A}=\frac{48}{7}$$

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$. By even symmetry, ${\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx={\displaystyle \frac{\sqrt{\pi }}{2}}$.

The slope is $8$ score points per additional hour. For a $0.5$-hour increase, the predicted change is $0.5\times 8=4$.

$f^{\mathrm{\prime }}(x)=3x^2-3a$, so $f^{\mathrm{\prime }}(2)=12-3a=0\Rightarrow a=4$. Check: $f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$, so $f^{\mathrm{\prime }\mathrm{\prime }}(2)=12>0$, confirming a local minimum.