What is the domain of the function f(x)=x^2+3x+5?

Mathematics

2 answers:

Ugo [173]3 years ago

8 0

Answer:

all real numbers

Step-by-step explanation:

Kitty [74]3 years ago

4 0

Answer:

The domain of this function is all real numbers

Step-by-step explanation:

Unless there is a domain restriction for the function, the domain of a function is all real numbers. This means that all we need to do is determine whether there is a domain restriction.

The things that cause a domain restriction are: an x in the denominator of a fraction like $f(x)=\frac{1}{x}$ or a radical like $f(x)=\sqrt{1-x}$

As there are no variables in the denominator or a radical, we can conclude that the domain of this function is all real numbers.

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What is the solution to the equation

Dafna11 [192]

$\sqrt{4t+5}=3-\sqrt{t+5}\\\\D:4t+5\geq0\ \wedge\ t+5\geq0\ \wedge\ \sqrt{t+5}\leq3\\\\t\geq-\dfrac{5}{4}\ \wedge\ t\geq-5\ \wedge\ t\leq6$

therefore
$D:x\in\left< -\dfrac{5}{4};\ 6\right>$

$\sqrt{4t+5}=3-\sqrt{t+5}\ \ \ |^2\\\\(\sqrt{4t+5})^2=(3-\sqrt{t+5})^2\ \ \ |use:(a-b)^2=a^2-2ab+b^2\\\\4t+5=3^2-2\cdot3\cdot\sqrt{t+5}+(\sqrt{t+5})^2\\\\4t+5=9-6\sqrt{t+5}+t+5\ \ \ \ |-t\\\\3t+5=14-6\sqrt{t+5}\ \ \ \ |-14\\\\3t-9=-6\sqrt{t+5}\ \ \ \ |change\ signs$
$9-3t=6\sqrt{t+5}\ \ \ \ |:3\\\\3-t=2\sqrt{t+5}\ \ \ \ |^2\\\\(3-t)^2=(2\sqrt{t+5})^2\\\\3^2-2\cdot3\cdot t+t^2=4(t+5)\\\\9-6t+t^2=4t+20\ \ \ |-4t-20\\\\t^2-10t-11=0\\\\t^2+t-11t-11=0\\\\t(t+1)-11(t+1)=0\\\\(t+1)(t-11)=0\iff t+1=0\ \vee\ t-11=0\\\\t=-1\in D\ \vee\ t=11\notin D$
Answer: t = -1.

7 0

3 years ago

Read 2 more answers

Find the value of x when 6-3x=5x-10x+4

tatuchka [14]

Answer:

x= -1

Step-by-step explanation:

6-3x=5x-10x+4

6-3x=-5x+4

-4 -4

2-3x=-5x