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4 years ago

What is the range of the function f(x) = (x + 3)2 + 4?

Mathematics

1 answer:

Tju [1.3M]4 years ago

7 0

Range of $f(x)=(x + 3)^2 + 4$ is [4,∞).

Step-by-step explanation:

Here we have , function $f(x)=(x + 3)^2 + 4$ , We need to find range of this function . Let's find out:

In the function $f(x)=(x + 3)^2 + 4$ , Let's focus on $(x+3)^2$ : It's a perfect square whose value will be always positive and greater then equal to zero i.e.

⇒ $(x+3)^2\geq 0$ , So minimum value of $(x+3)^2$ is 0 at x=-3 . ∴

⇒ $f(x)=(x + 3)^2 + 4$

⇒ $f(-3)=(-3+ 3)^2 + 4$

⇒ $f(-3)= 4$

Minimum value of $f(x)=(x + 3)^2 + 4$ is 4 , And maximum value of $f(x)=(x + 3)^2 + 4$ can be ∞ as value of x can be increased . Therefore , Range of $f(x)=(x + 3)^2 + 4$ is [4,∞).

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Joan, Kylie, Lillian, and Miriam all celebrate their birthdays today. Joan is 2 years younger than Kylie, Kylie is 3 years older

snow_tiger [21]

Answer:

d) 54

Step-by-step explanation:

Let's name x the age of Kylie.

Then we have:

K = x

Joan is 2 years younger than Kylie ⇒ J = x - 2

Kylie is 3 years older than Lilian ⇒ L = x - 3

Miriam is one year older than Joan ⇒M = (x - 2) + 1 = x - 1

If we sum up all ages we get: x + x -2 + x - 3 + x - 1 = 4x - 6

Therefore, the sum of all ages is 4x - 6.

Let's make x = 14 ⇒ 4x - 6 = 4(14) - 6 = 50

Let's make x = 15 ⇒4x - 6 = 4(15) - 6 = 54

Let's make x = 16 ⇒4x - 6 = 4(16) - 6 = 58

If we substitute for smaller x, we'll get values smaller than 50 and if we substitute for x >16 we'll get values greater than 58.

Looking at the options given by the problem, we can see that the only one that fits into the equation is d) 54.

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3 years ago

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Kryger [21]

Answer:

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Step-by-step explanation:

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3 years ago

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Let a and b be real constants such that \[x^4 ax^3 3x^2 bx 1 \ge 0\]for all real numbers x. Find the largest possible value of a

tatiyna

Answer:

Infinity

Step-by-step explanation:

Since $x^4 ax^3 3x^2 bx 1 \ge 0$ for all real numbers x, this property is also true for x=1, which tells us that

$1^4 a1^3 3\cdot1^2 b\cdot 1=3ab\ge0$

On the other hand, note that for all real numbers x, it holds that

$x^4x^3x^2x\ge 0$

Therefore, if

$3ab\geq0$

we have tat

$3ab x^{4}x^{3}x^{2}x^{1}1=x^4ax^3 3x^2bx1\ge0$

The last reasoning tells us that the property $x^4 ax^3 3x^2 bx 1 \ge 0$ holds for all real numbers x if an only if $ab\ge0$

Therefore, we can choose arbitrary constants a and b as long as

$ab\ge0$

We can choose a and b such that both positive, both negative or one of the two constants is equal two zero. In the first two cases

$a^2b^2$

can get as big as we want, depending on the constants, and we are done.

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3 years ago

With of the following is true about the expression √3*√12

scoundrel [369]

Answer:

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Step-by-step explanation:

When you multiply the square root of 3 and 12 together you get the square root of 36 which equals:

6 is your answer.

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4 years ago

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-5 Which expression is equivalent to (2^3)^-5?

kompoz [17]

$\huge\text{Hey there!}$

$\large\boxed{\mathsf{(2^3)^{-5}}}}\\\\\mathsf{2^3}\\\mathsf{= 2*2*2}\\\mathsf{= 4*2}\\\mathsf{= 8}\\\\\large\boxed{\mathsf{8^{-5}}}\\\mathsf{\rightarrow \mathsf{\dfrac{1}{8*8*8*8*8}}}\\\\\large\boxed{\rightarrow \mathsf{\dfrac{1}{6 4*6 4* 8}}}\\\\\large\boxed{\mathsf{\rightarrow \dfrac{1}{4,096*8}}}$

$\large\boxed{\mathsf{\rightarrow \dfrac{1}{32,768}}}\\\\\huge\textbf{Thus, your answer is: \boxed{\mathsf{\dfrac{1}{32,768}}}}}}$

$\huge\textbf{or \boxed{\mathsf{8^{-5}}}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

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