Question

for problem 7 & 8, the given limits represent an f(c) for a function f(x) and a number c. find f and c.

1. $lim_{delta x

ightarrow0}\frac{5 - 3(1+delta x)-2}{delta x}$

1. $lim_{delta x

ightarrow0}\frac{(-2+delta x)^3 + 8}{delta x}$

Explanation:

Step1: Recall the definition of the derivative

The derivative of a function $f(x)$ at $x = c$ is defined as $f^{\prime}(c)=\lim_{\Delta x
ightarrow0}\frac{f(c + \Delta x)-f(c)}{\Delta x}$.

Step2: Solve problem 7

We have $\lim_{\Delta x
ightarrow0}\frac{[5 - 3(1+\Delta x)]-2}{\Delta x}$. First, simplify the numerator: $5-3(1 + \Delta x)-2=5-3 - 3\Delta x-2=-3\Delta x$. So the limit is $\lim_{\Delta x
ightarrow0}\frac{-3\Delta x}{\Delta x}=-3$. Comparing with the derivative - definition $f^{\prime}(c)=\lim_{\Delta x
ightarrow0}\frac{f(c+\Delta x)-f(c)}{\Delta x}$, we can let $c = 1$ and $f(x)=5-3x$. Because $f(1)=5 - 3\times1 = 2$ and $f(1+\Delta x)=5-3(1+\Delta x)$.

Step3: Solve problem 8

We have $\lim_{\Delta x
ightarrow0}\frac{(-2+\Delta x)^{3}+8}{\Delta x}$. Expand $(-2+\Delta x)^{3}$ using the formula $(a + b)^{3}=a^{3}+3a^{2}b + 3ab^{2}+b^{3}$, where $a=-2$ and $b = \Delta x$. So $(-2+\Delta x)^{3}=(-2)^{3}+3\times(-2)^{2}\Delta x+3\times(-2)(\Delta x)^{2}+(\Delta x)^{3}=-8 + 12\Delta x-6(\Delta x)^{2}+(\Delta x)^{3}$. Then $(-2+\Delta x)^{3}+8=12\Delta x-6(\Delta x)^{2}+(\Delta x)^{3}$. The limit becomes $\lim_{\Delta x
ightarrow0}\frac{12\Delta x-6(\Delta x)^{2}+(\Delta x)^{3}}{\Delta x}=\lim_{\Delta x
ightarrow0}(12-6\Delta x+\Delta x^{2}) = 12$. Comparing with the derivative - definition, we can let $c=-2$ and $f(x)=x^{3}$, since $f(-2)=(-2)^{3}=-8$ and $f(-2+\Delta x)=(-2+\Delta x)^{3}$.

Answer:

1. $f(x)=5 - 3x$, $c = 1$
2. $f(x)=x^{3}$, $c=-2$