Question

KK bantuin pertanyaanku​

Answer 1

Verified answer

Soal 1

[tex] \small\begin{aligned} \lim_{x\to\infty} \frac{ 4x^2+3x-10}{ 7-11x+2x^2} &= \lim_{x\to\infty} \frac{ 4x^2+3x-10}{ 7-11x+2x^2} \cdot \frac{ \frac{1}{x^2}}{ \frac{ 1}{x^2}} \\ &= \lim_{x\to\infty} \frac{ 4+\frac{ 3}{x}- \frac{ 10}{x^2}}{ \frac{ 7}{x^2}- \frac{11}{x}+2} \\ &= \frac{ 4+0-0}{ 0-0+2} \\ &= \frac 42 \\ &= 2\end{aligned} [/tex]

Soal 2

Tentukan turunan [tex] f(x):[/tex]

[tex] \begin{aligned} f(x) &= 2x^3-9x^2-24x \\ &\downarrow \\ f'(x) &= 6x^2-18x-24 \end{aligned} [/tex]

Syarat stasioner:

[tex]\begin{aligned} f'(x) &= 0 \\ 6x^2-18x-24 &= 0 \\ x^2-3x-4 &= 0 \\ (x+1)(x-4) &= 0 \\ x=-1 \text{ atau } x &= 4\end{aligned} [/tex]

Jadi, Nilai stasion [tex] f(x) [/tex] tercapai saat nilai [tex] x=-1 \text{ atau } x=4.[/tex]

Soal 3

Tentukan turunan [tex]f(x): [/tex]

[tex] \begin{aligned} f(x) &= \frac45x^5+\frac23x^3+x-2 \\ &\downarrow \\ f'(x) &= 4x^4+2x^2+1\end{aligned} [/tex]

Tentukan nilai [tex] f'(-2)[/tex] dengan substitusi [tex] x=-2 [/tex] ke hasil turunan:

[tex] \begin{aligned} f'(2) &= 4(-2)^4+2(-2)^2+1 \\ &= 4(16)+2(4)+1 \\ &= 64+8+1 \\ &= 73 \end{aligned} [/tex]

Jadi, nilai dari [tex] f(-2)[/tex] adalag [tex]73. [/tex]