Y=6cosx^2 (derivative)

Answers

Answer 1

Answer: dy/dx = d/dx 6cosx^2
= 12cosx d/dx cosx
= 12cosx (-sinx)
= -12cosxsinx

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Graph and find the x-intercept, y-intercept, domain, range, and horizontal asymptote of the function y = 4x

Answers

The function $y=4x$ is a line. As such, its domain and range is the whole real number set $\mathbb{R}$ , and it has no horizontal nor vertical asymptotes.

Moreover, there's no constant term, so it's x and y intercept is the origin (0,0).

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought in to the time repairs are completed. A random sample of 12 repair records showed the following repair times (in days): 5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.(a) H0: μ ≤ 5 days versus H1: μ > 5 days. At α = .05, choose the right option. Reject H0 if tcalc > 1.7960
Reject H0 if tcalc < 1.7960

b. Calculate the Test statistic.

c-1. The null hypothesis should be rejected.
i. TRUE
ii. FALSE

c-2. The average repair time is longer than 5 days.
i. TRUE
ii. FALSE

c-3 At α = .05 is the goal being met?
i. TRUE
ii. FALSE

Answers

Answer:

a) Reject H0 if tcalc > 1.7960

b) $t=(5.42-5)/((1.16)/(√(12)))=1.239$

c-1) ii. FALSE

c-2) ii.FALSE

c-3)i. TRUE

Step-by-step explanation:

1) Data given and notation

$\bar X=5.42$ represent the mean time for the sample

$s=1.16$ represent the sample standard deviation for the sample

$n=12$ sample size

$\mu_o =5$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

a) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 5 days, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5$

Alternative hypothesis: $\mu > 5$

We don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=(\bar X-\mu_o)/((s)/(√(n)))$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Rejection zone

On this case we need a critical value that accumulates 0.05 of the area on the right tail. The degrees of freedom are given by 11. And we can use the following excel code to find the critical value : "T.INV(1-0.95,11)" and the critical value would be given by $t_(\alpha/2)=1.795$ .

And the rejection zone is given by:

Reject H0 if tcalc > 1.7960

b) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=(5.42-5)/((1.16)/(√(12)))=1.239$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=12-1=11$

Since is a one side test the p value would be:

$p_v =P(t_((11))>1.239)=0.121$

c-1. The null hypothesis should be rejected.

ii. FALSE

c-2. The average repair time is longer than 5 days.

ii. FALSE

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, and the true mean is not significantly higher than 5.

c-3 At α = .05 is the goal being met?

i. TRUE

We fail to reject the null hypothesis so then the goal is met.

Let h be the function given by h(x) =x+x-2
²-1
We will investigate the behavior of
both the numerator and denominator of h(x) near the point where x = 1. Let
f(x)= x³ + x -2 and g(x)=x²-1. Find the local linearizations of f and g at a = 1,
and call these functions Lf(x) and Lg(x), respectively.
Lf(x) =
L₂(x) =
Explain why h(x) ≈
Lf(x)
Lg(x)
for a near a = 1.

Answers

Final answer:

The local linearizations of f(x) and g(x) at a = 1 are Lf(x) = 4x - 5 and Lg(x) = 2x - 2 respectively. The function h(x) ≈ Lf(x)/Lg(x) because the local linearizations provide a good approximation of the numerator and denominator of h(x) near x = 1.

Explanation:

The local linearization of a function at a given point is an approximation of the function using a linear equation. To find the local linearization of a function f at a = 1, we need to find the slope of the tangent line at a = 1, which is equivalent to finding the derivative of f at x = 1. By taking the derivative of f(x) = x³ + x - 2, we get f'(x) = 3x² + 1. Evaluating f'(1), we find that the slope of the tangent line at a = 1 is 4. Therefore, the local linearization of f at a = 1, denoted as Lf(x), is given by Lf(x) = f(a) + f'(a)(x - a), which becomes Lf(x) = -1 + 4(x - 1) = 4x - 5.

Similarly, to find the local linearization of g(x) = x² - 1 at a = 1, we need to find the slope of the tangent line at a = 1. The derivative of g(x) is g'(x) = 2x. Evaluating g'(1), we find that the slope of the tangent line at a = 1 is 2. Therefore, the local linearization of g at a = 1, denoted as Lg(x), is given by Lg(x) = g(a) + g'(a)(x - a), which becomes Lg(x) = 0 + 2(x - 1) = 2x - 2.

When investigating the behavior of the function h(x) = (f(x))/(g(x)) near the point x = 1, we can approximate h(x) using the local linearizations of f and g at a = 1. Near the point a = 1, h(x) ≈ Lf(x)/Lg(x) because Lf(x) and Lg(x) provide a good approximation of the numerator and denominator, respectively, of h(x). This approximation holds as long as x is close to 1.

Learn more about local linearizations of functions here:

brainly.com/question/34258619

Help---I dont understand!! :(

Answers

Answer:

it should be c

Step-by-step explanation:

on edg hope this helps

Does the residual plot suggest a linear relationship?

Answers

Answer:

Step-by-step explanation:

Find the critical points, domain endpoints, and local extreme values for the functiony=x^2/5(x+3)

a. What is/are the critical point(s) and domain endpoint(s) where f' is undefined?
b. What is/are the critical point(s) and domain endpoint(s) where f' is 0?
c. From the critical point(s) and domain endpoint(s), what is/are the points corresponding to local maxima?
d. From the critical point(s) and domain endpoint(s), what is/are the points corresponding to local minima?

Answers

Answer:

a) $x = -3$ , b) $x = 0$ , $x = -6$ , c) $x = 0$ , d) $x = -6$

Step-by-step explanation:

a) Let derive the function:

$f'(x) = (10\cdot x \cdot (x+3)-5\cdot x^(2))/(25\cdot (x+3)^(2))$

$f'(x)$ is undefined when denominator equates to zero. The critical point is:

$x = -3$

b) $f'(x) = 0$ when numerator equates to zero. That is:

$10\cdot x \cdot (x+3) - 5\cdot x^(2) = 0$

$10\cdot x^(2)+30\cdot x -5\cdot x^(2) = 0$

$5\cdot x^(2) + 30\cdot x = 0$

$5\cdot x \cdot (x+6) = 0$

This equation shows two critical points:

$x = 0$ , $x = -6$

c) The critical points found in point b) and the existence of a discontinuity in point a) lead to the conclusion of the existence local minima and maxima. By plotting the function, it is evident that $x = 0$ corresponds to a local maximum. (See Attachment)

d) By plotting the function, it is evident that $x = -6$ corresponds to a local minimum. (See Attachment)

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