Question

lim(x→∞) (1 + 2x² - x³)/(3x³ - 2x + 1)
f. use first principle f(x)=lim(h→0) (f(x + h)-f(x))/h to show that the derivative of f(x)=x² - 2x
g. determine the value of x that f(x)=2x² - 8 increasing.
h. consider the curve given by the function f(x)=x³ - 3x² + 2x. determine the equation of the tangent line to the curve at the point where x = 1.

Explanation:

Step1: Solve $\lim_{x

ightarrow\infty}\frac{1 + 2x^{2}-x^{3}}{3x^{3}-2x + 1}$
Divide numerator and denominator by $x^{3}$:

$$\lim_{x ightarrow\infty}\frac{\frac{1}{x^{3}}+\frac{2}{x}- 1}{3-\frac{2}{x^{2}}+\frac{1}{x^{3}}}$$

As $x
ightarrow\infty$, $\frac{1}{x^{n}}
ightarrow0$ for $n>0$. So the limit is $\frac{0 + 0-1}{3-0 + 0}=-\frac{1}{3}$.

Step2: Use first - principle for $f(x)=x^{2}-2x$

First, find $f(x + h)$:
$f(x + h)=(x + h)^{2}-2(x + h)=x^{2}+2xh+h^{2}-2x-2h$
Then, $f(x + h)-f(x)=x^{2}+2xh+h^{2}-2x-2h-(x^{2}-2x)=2xh+h^{2}-2h$
So, $f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h
ightarrow0}\frac{2xh+h^{2}-2h}{h}=\lim_{h
ightarrow0}(2x + h-2)=2x-2$

Step3: Determine where $f(x)=2x^{2}-8$ is increasing

Find the derivative $f^{\prime}(x)=4x$.
A function is increasing when $f^{\prime}(x)>0$. So, $4x>0$, which gives $x>0$.

Step4: Find the equation of the tangent line for $f(x)=x^{3}-3x^{2}+2x$ at $x = 1$

First, find $f(1)$: $f(1)=1^{3}-3\times1^{2}+2\times1=1 - 3+2=0$
Then, find the derivative $f^{\prime}(x)=3x^{2}-6x + 2$.
$f^{\prime}(1)=3\times1^{2}-6\times1 + 2=3-6 + 2=-1$
The equation of the tangent line using the point - slope form $y - y_{1}=m(x - x_{1})$ (where $(x_{1},y_{1})=(1,0)$ and $m=-1$) is $y-0=-1(x - 1)$, or $y=-x + 1$.

Answer:

• $\lim_{x

ightarrow\infty}\frac{1 + 2x^{2}-x^{3}}{3x^{3}-2x + 1}=-\frac{1}{3}$

• $f^{\prime}(x)$ for $f(x)=x^{2}-2x$ using first - principle is $2x - 2$
• $f(x)=2x^{2}-8$ is increasing for $x>0$
• The equation of the tangent line for $f(x)=x^{3}-3x^{2}+2x$ at $x = 1$ is $y=-x + 1$