Ask question

Ask question

All categories

2 years ago

11

The graph of f ′ (x), the derivative of f of x, is continuous for all x and consists of five line segments as shown below. Given

f (–3) = 6, find the absolute maximum value of f (x) over the interval [–3, 0].
Graph of line segments increasing from x equals negative 4 to x equals negative 3, decreasing from x equals negative 3 to x equals 0, increasing from x equals 0 to x equals 3, constant from x equals 2 to x equals 4 and decreases from x equals 4 to x equals 5. x intercepts at x equals negative 4, x equals 0, x equals 5.

3
4.5
6
10.5

1 answer:

myrzilka [38]2 years ago

4 0

Answer

10.5

Step-by-step explanation:

The other answer is correct until solving for C. f(-3) = 6, not -6. So C = 21/2. This makes f(0) = 10.5. Local Max should occur at x intercepts on the f'(x) graph.

You might be interested in

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10

zvonat [6]

The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

$FV=P\left(1-\dfrac{r}{100}\right)^n$

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is $n_1$ . Thus, by the above formula for the first car,

$0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}$

Take log both the sides as,

$\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85$

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is $n_2$ . Thus, by the above formula for the second car,

$0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}$

Take log both the sides as,

$\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14$

The difference in the ages of the two cars is,

$d=4.85-3.14\\d=1.71\rm years$

Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0

2 years ago

The gcf of 135 225 270

rodikova [14]

The greatest common factor of 135 225 and 270 is 135/225 = 3/5

hope you found this helpful ;)

5 0

3 years ago

Use the value of the first integral I to evaluate the two given integrals. I = integral^3_0 (x^3 - 4x)dx = 9/4 A. integral^3_0 (

Troyanec [42]

Answer:

A) -9/2

B) 9/4

C) -9/2, same as A)

Step-by-step explanation:

We are given that $I=\int_0^3 x^3-4x dx=9/4$ . We use the properties of integrals to write the new integrals in terms of I.

A) $\int_0^3 8x-2x^3 dx=\int_0^3 -2(x^3-4x) dx=-2\int_0^3 x^3-4x dx=-2I=-9/2$ . We have used that ∫cf dx=c∫f dx.

B) $\int_3^0 4x - x^3 dx=-\int_0^3 (4x-x^3) dx=\int_0^3-(4x-x^3) dx=\int_0^3 x^3-4x dx=I=9/4$ . Here we used that reversing the limits of integration changes the sign of the integral.

C) It's the same integral in A)

4 0

2 years ago

If onelb of shrimp is 1 5/8 and it costs 16.99 how much do you need to get 5 lb of shrimp

lyudmila [28]

Answer:

If one is 1 5/8 and its cost is 16.99 and you need 5lb.

Step-by-step explanation:Than first you will take the 16.99 x 5 =84.95 dollars. Take that 1 5/8 x 5 = 8 1/8

So it would take 8 1/8 and 84.95$

7 0

2 years ago

Which of the following statements are true? Select all that apply.

Tom [10]

Answer: The statements in options 2 and 4 are true.

Explanation:

The LCM of two numbers is the smallest number that is multiple of both numbers.

The factor of 8 and 12 are,

$8=2\times 2\times 2$

$12=2\times 2\times 3$

Since 2 and 2 are common in both but 2 and 3 are not same, therefore the L.C.M. of 8 and 12 is,

$L.C.M.(8,12)=2\times 2\times 2\times 3=24$

Therefore the statement in option 1 is incorrect.

The factor of 6 and 9 are,

$6=2\times 3$

$9=3\times 3$

Since 3 is common in both but 2 and 3 are not same, therefore the L.C.M. of 6 and 9 is,

$L.C.M.(6,9)=2\times 3\times 3=18$

Therefore the statement in option 2 is correct.

The factor of 11 and 4 are,

$11=1\times 11$

$4=2\times 2$

Since all factor are different, therefore the L.C.M. of 11 and 4 is,

$L.C.M.(11,4)=1\times 11\times 2\times 2=44$

Therefore the statement in option 3 is incorrect.

The factor of 9 and 10 are,

$9=1\times 9$

$10=2\times 5$

Since all factor are different, therefore the L.C.M. of 9 and 10 is,

$L.C.M.(9,10)=1\times 9\times 2\times 5=90$

Therefore the statement in option 4 is correct.

4 0

2 years ago