Ask question

Ask question

All categories

3 years ago

5

Let the function f be continuous and differentiable for all x. Suppose you are given that f(−1)=−3, and that f'(x) for all value

s of x. Use the Mean Value Theorem to determine the largest possible value of f(5).

1 answer:

Monica [59]3 years ago

8 0

Answer:

Step-by-step explanation:

By MVT,

$f'(c) = \frac{f(5)-f(-1)}{5-1}$

$4f'(c)= f(5) + 3\rightarrow f(5) = 4f'(c) - 3$

Moreover, $f'(x)\leq 4~~~~~~ \forall x,$

$f(5) \leq 16 - 3 =13$

Hence the largest possible value of f(5) is 13

You might be interested in

A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

4 years ago

An explosion causes debris to rise vertically with an initial velocity of 160 feet per second. What is the speed of debris when

lisov135 [29]

The question is asking what v_final is, given that v_initial is at 300 feet. and v_initial is at 0 feet.
We know there will be a constant downward acceleration of 32.15 ft/s^2.
Use the following equation:
v_final^2 = v_initial^2 + 2ah
v_final^2 = (160 ft/s)^2 + 2(-32.15 ft/s^2)(300 ft) = 6310 ft^2/s^2
v_final = (6310 ft^2/s^2)^1/2 = 79.4 ft/s.

5 0

3 years ago

1. What is the solution to the system?

Mariulka [41]

Answer:

D.(1, -1, 2) hope this helps

7 0

3 years ago

Read 2 more answers

I NEED HELP PLEASE ASAP PLEASE IM BEGGING YOU PLEASE!!!!!!!!!!!

Evgesh-ka [11]

Where’s the picture?

6 0

3 years ago

Read 2 more answers

HELP!!! WILL BE MARK AS A BRAINLISTTTT!!! Please

jonny [76]

I believe the answer is b

8 0

3 years ago