2 years ago

Which point lies on the line described by the equation below? y+4=4(x-3

1 answer:

3 0

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies \begin{array}{llll} y+4=4(x-3)\implies y-(\stackrel{y_1}{-4})=\stackrel{m}{4}(x-\stackrel{x_1}{3})\\\\ ~\hfill (\stackrel{x_1}{3}~~,~~\stackrel{y_1}{-4}) \end{array}$

You might be interested in

Determine the values of the constants B and C so that the function given below is differentiable.

laila [671]

For the function to be differentiable, its derivative has to exist everywhere, which means the derivative itself must be continuous. Differentiating gives

$f'(x)=\begin{cases}24x^2&\text{for }x1\end{cases}$

The question mark is a placeholder, and if the derivative is to be continuous, then the question mark will have the same value as the limit as $x\to1$ from either side.

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}24x^2=24$
$\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}B=B$

So the derivative will be continuous as long as $B=24$

For the function to be differentiable everywhere, we need to require that $f(x)$ is itself continuous, which means the following limits should be the same:

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}8x^3=8$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}Bx+C=24+C$

$24+C=8\implies C=-16$

So, the function should be

$f(x)=\begin{cases}8x^3&\text{for }x\le1\\24x-16&\text{for }x>1\end{cases}$

with derivative

$f'(x)=\begin{cases}24x^2&\text{for }x$

5 0

4 years ago

Let ff be the function given by f(x)=e2x−1 x f(x)=e2x−1 x. Which of the following equations expresses the property that f(x)f(x)

navik [9.2K]

Answer:

Step-by-step explanation:

$\lim_{x \to \ 0} \frac{e^{2x}-1 }{x} \\= \lim_{x \to \0} \frac{e^{2x}-1 }{2 x} *2\\\rightarrow2 log e\\\rightarrow 2$