Help me plez and show work

Solve the inequality and write down the solution in interval notation x(x-3)(x+1)<0

Ratling [72]

Since it's so nicely grouped, we can work with it! For the equation to equal 0, x=0, 3, or -1 (since x-3 and x+1 equal 0 when plugged in with 3 and -1 respectively). All we have to do is plug in numbers before, between, and after these numbers and apply it to the rest of them. Since -1 is the smallest number of the group, we can plug in a number below that (for this example, -5) and plug it in to get -8*-5*-4= something negative since it contains an odd number of negative numbers. Therefore, anything less than 1 is negative. For between -1 and 0, we get x=-0.5 equals -0.5*-3.5*0.5=something positive (since it has an even amount of negative numbers), proving that everything between -1 and 0 here is positive. For something between 0 and 3, we can plug 1 in to get 1*-2*2= something negative. Do you see a pattern here? It's negative, then positive, etc.. Therefore, if the number is greater than 3 it is positive. Reviewing a bit, we can see that (-inf, -1) is negative as well as (0,3), making the interval notation (-inf, -1) U (0, 3) since when you plug -1, 0, and 3 in it is 0, not less than 0!

8 0

3 years ago

PLEASE HELP A student at a university makes money by buying and selling used cars. Charles bought a used car and later sold it f

Fed [463]

Answer: Charles bought a used car and later sold it for a 20% profit. If he sold it for $4680, how much did Charles pay for the car?

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Let the price he paid equal "x".

EQUATION:

x + 0.20x = 4680

1.2x = 4680

x = $3900 (price he paid for the car.

5 0

3 years ago

PLEASE ANSWER ASAP!!

Rufina [12.5K]

Answer:

The instantaneous rate of increase of f(x) at $x_{0} = 3$ is 3.
One possible equation of this line is y = 3x - 16.
The line is tangent to the graph of y = f(x). The slope of the line is the same as the instantaneous rate of increase of f(x) at $x_{0} = 3$ .

Step-by-step explanation:

$\begin{aligned}&\lim_{h\to 0}{\frac{f(3+h)-f(h)}{h}}\\ =&\lim_{h\to 0}\frac{1}{h} \cdot [(3^{3} + 3 \times 3^{2}h +3\times 3h^{2} + h^{3})- 4(3^{2} + 2\times 3h + h^{2}) + 2 \\[-1em] &\phantom{\lim_{h\to 0}\frac{1}{h}\cdot []} -(3^{3} -4\times 3^{2} + 2)]\\ = &\lim_{h \to 0}\frac{h^{3} + 5h^{2} +3h}{h}\\=& \lim_{h \to 0}{h^{2}} + \lim_{h \to 0}{5h} + \lim_{h\to 0}{3}\\ =& 3\end{aligned}$ .

$f(3) = 3^{3} - 4\times 3^{2} + 2 =-7$ .

In other words, the graph of y = f(x) passes through the point (3, -7) where $x = 3$ .

The point-slope form of a line in a cartesian plane is:

$y - y_0 = m (x - x_0)$ .

For this line,

$(3, -7)$ is the point on the line, while

$3$ is the slope of the line.

The equation of this line will thus be

$y - (-7) = 3(x - 3)$ .

That's equivalent to

$y = 3x - 16$ .

Refer to the diagram attached. The line touches the graph of y = f(x) at x = 3 without crossing it. The line here is thus a tangent to the graph of y = f(x) at x = 3. The slope of the line represents the instantaneous rate of increase of f(x) at $x_{0} = 3$ .