All categories

3 years ago

Consider the function given by the graph.

Mathematics

2 answers:

alukav5142 [94]3 years ago

3 0

Answer:

f(-2 ) =2

f(0) = 3

f(4) = -1

Step-by-step explanation:

Clearly by looking at the graph of the given function f(x) we could observe that the function f(x) is increasing in the interval (-∞,0] and then decreasing in the interval (0,∞).

Also, the function is discontinuous at x=0.

( Since, there is a break in a graph and also the left and right hand limit of the function are not equal at x=0)

Hence, from the graph we could directly say that:

f(-2 ) =2

f(0) = 3

f(4) = -1

dlinn [17]3 years ago

3 0

f(–2 ) = 2
f(0) = 3
f(4) = - 1

You might be interested in

Verify the identity (tan x + 1)^2 + (tan x-1)^2= 2 sec^2 x

Elina [12.6K]

$(\tan x+1)^2+(\tan x-1)^2=2\sec^2x\\\\\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\\\\L_s=\left(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}\right)^2\\\\=\left(\dfrac{\sin x+\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x-\cos x}{\cos x}\right)^2\\\\=\dfrac{(\sin x+\cos x)^2}{\cos^2x}+\dfrac{(\sin x-\cos x)^2}{\cos^2x}\\\\\text{use}\ (a\pm b)^2=a^2\pm2ab+b^2$

$=\dfrac{\sin^2x+2\sin x\cos x+\cos^2}{\cos^2x}+\dfrac{\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{\sin^2x+2\sin x\cos x+\cos^2+\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{2\sin^2x+2\cos^2x}{\cos^2x}=\dfrac{2(\sin^2x+\cos^2x)}{\cos^2x}\\\\\text{use}\ \sin^2x+\cos^2x=1\\\\=\dfrac{2(1)}{\cos^2x}=2\cdot\dfrac{1}{\cos^2x}=2\left(\dfrac{1}{\cos x}\right)^2\\\\\text{use}\ \sec x=\dfrac{1}{\cos x}\\\\=2(\sec^2x)=2\sec^2x=R_s\\\\L_s=R_s\Rightarrow The\ identity$

4 0

3 years ago

1 What is the value of this expression? 2 +4² x 4 - 12

____ [38]

Tell Me If Somethings wrong with my answer!

7 0

3 years ago

Read 2 more answers

Someone please help me with this

amid [387]

Simliar-AA is your answer

7 0

3 years ago

Factor completely 2x^3 + 6x^2 + 10x + 30

andrezito [222]

Answer:

The factored expression is 2(x² + 5)(x + 3).

Step-by-step explanation:

Hey there!

We can use a factoring technique referred to as "grouping" to solve this problem.

Grouping is used for polynomials with four terms as a quick and easy factoring method to remove the GCF and get down to the initial terms that create the expression/function.

Grouping works in the following matter:

Given equation: ax³ + bx² + cx + d
Group a & b, c & d: (ax³ + bx²) + (cx + d)
Pull GCFs and factors

Let's apply these steps to the given equation.

Given equation: 2x³ + 6x² + 10x + 30
Group a & b, c & d: (2x³ + 6x²) + (10x + 30)
Pull GCFs and factors: 2x²(x + 3) + 10(x + 3)

As you'll see, we have a common term with both sides of the expression. This term, (x + 3), is a valuable asset to the factoring process. This is one of the factors for our expression.

Now, we use our GCFs to create another factor.

List GCFs: 2x², 10
Create a term: (2x² + 10)

Finally, we'll need to simplify this one by taking another GCF, 2.

Pull GCF: 2(x² + 5)

Now that we have this term, we need to understand that this could also be factored further using imaginary numbers, but it is also acceptable to leave it in this form.

Therefore, we have our final factors: 2(x² + 5) and (x + 3).

However, when we factor, we place all of our terms together. This leaves us with the final answer: 2(x² + 5)(x + 3).

7 0

3 years ago

Find the y intercepts of 3x^2+24x-51. quadratic formula

julia-pushkina [17]