Classify the percentages based on their value.

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$

$f'(x)=12x^2+26x-12$

$f''(x)=24x+26$

So if f''(a) is positive then we have a minimum at x=a.

If f''(a) is negative then we have a maximum at x=a.

Rounding to nearest tenths here: x=-2.6 and x=.4

Let's see what f'' gives us at both of these x's.

$24(-2.6)+25$

$-37.5$

So we have a maximum at x=-2.6.

$24(.4)+25$

$9.6+25$

$34.6$

So we have a minimum at x=.4.

Now let's find the corresponding y-value for our relative minimum point since that would complete your question.

We are going to use the equation that relates x and y.

I'm going to use 0.3909838 instead of .4 just so we can be closer to the correct y value.

$y=4(0.3909838)^3+13(0.3909838)^2-12(0.3909838)-56$

I'm shoving this into a calculator:

$y=-58.4654411$

So the approximate relative minimum is (0.4,-58.5).

If you graph $y=4x^3+13x^2-12x-56$ you should see the graph taking a dip at this point.

3 0

4 years ago

Plz help will choose brainliest provide an explanation.

nika2105 [10]

Answer:

I wanna say b

Step-by-step explanation:

I'm gonna say b because u would think since they've paid $13, and it sells for $22, u would add that on the price, which in my opinion would be x + 13 = 22

7 0

3 years ago

In a class of 7 there are 2 students who forgot their lunch. If the teacher chooses 2 students what is the probability that both

Reika [66]

It is about a 28% chance.

4 0

3 years ago

Which expression is equivalent to (3x² + 3x - 3) - (6x² - 4x - 2)?

Marat540 [252]

Answer:

A. $12x^{4} -4x^{3} +3x^{2} +3x+20$

Step-by-step explanation:

For this problem, we have to combine like terms.

Let's look at the equation:

$(12x^{4} -4x^{3} +7x+6)+(3x^{2} -4x+14)$

$12x^{4$ , $-4x^{3}$ , $3x^{2}$ do not have any other like terms, so we keep all in the final answer.

$7x$ and $-4x$ are like terms, so we combine them. $7x-4x=3x$

14 and 6 are like terms, so we combine them. 14+6 = 20

Let's put everything into the final equation:

$12x^{4} -4x^{3} +3x^{2} +3x+20$

So, the final answer is A. $12x^{4} -4x^{3} +3x^{2} +3x+20$

Hope this helps! If you have any questions about my work, leave them in the comments below!

5 0

2 years ago

Need help please .The ares of the sector shown below is 59 yd^2 find the length of the radius. Round to the nearest whole number

Vinvika [58]

$\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360} ~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ A=59\\ \theta =40 \end{cases}\implies 59=\cfrac{(40)\pi r^2}{360}\implies 59(360)=40\pi r^2 \\\\\\ \cfrac{59(360)}{40\pi }=r^2\implies \cfrac{531}{\pi }=r^2\implies \sqrt{\cfrac{531}{\pi }}=r\implies 13\approx r$

3 0

2 years ago