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HACTEHA [7]
3 years ago
6

Use the functions f(x)=2x and g(x)= x^2+1 to find the value of each expression.

Mathematics
1 answer:
musickatia [10]3 years ago
8 0

Answer:

First, you learned (back in grammar school) that you can add, subtract, multiply, and divide numbers. Then you learned that you can add, subtract, multiply, and divide polynomials. Now you will learn that you can also add, subtract, multiply, and divide functions. Performing these operations on functions is no more complicated than the notation itself. For instance, when they give you the formulas for two functions and tell you to find the sum, all they're telling you to do is add the two formulas. There's nothing more to this topic than that, other than perhaps some simplification of the expressions involved.

Step-by-step explanation:

To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.

(f + g)(x) = f (x) + g(x)

= [3x + 2] + [4 – 5x]

= 3x + 2 + 4 – 5x

= 3x – 5x + 2 + 4

= –2x + 6

(f – g)(x) = f (x) – g(x)

= [3x + 2] – [4 – 5x]

= 3x + 2 – 4 + 5x

= 3x + 5x + 2 – 4

= 8x – 2

(f  × g)(x) = [f (x)][g(x)]

= (3x + 2)(4 – 5x)

= 12x + 8 – 15x2 – 10x

= –15x2 + 2x + 8

\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(  

g

f

​  

)(x)=  

g(x)

f(x)

​  

 

= \small{\dfrac{3x+2}{4-5x}}=  

4−5x

3x+2

​  

 

My answer is the neat listing of each of my results, clearly labelled as to which is which.

( f + g ) (x) = –2x + 6

( f – g ) (x) = 8x – 2

( f  × g ) (x) = –15x2 + 2x + 8

\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}(  

g

f

​  

)(x)=  

4−5x

3x+2

​

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2 years ago
Find the directional derivative of f(x,y,z)=z3−x2yf(x,y,z)=z3−x2y at the point (−5,5,2)(−5,5,2) in the direction of the vector v
olga_2 [115]

We are given

f=z^3 -x^2y

Firstly, we can find gradient

so, we will find partial derivatives

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f_y=-x^2

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now, we can plug point (-5,5,2)

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so, gradient will be

gradf=(50,-25,12)

now, we are given that

it is in direction of v=⟨−3,2,−4⟩

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|v|=\sqrt{29}

now, we can find unit vector

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now, we can plug values

dir=(50,-25,12) \cdot (\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })

dir=(-\frac{150}{\sqrt{29} } - \frac{50}{\sqrt{29} } - \frac{48}{\sqrt{29} })

dir=-\frac{248\sqrt{29}}{29}.............Answer



7 0
3 years ago
Read 2 more answers
U(8, 9) and V(2, 5) are the endpoints of a line segment. What is the midpoint M of that line segment?
Pavel [41]

Answer:

The midpoint M is (5,7)

Step-by-step explanation:

m = ( \frac{x1 + x2}{2} , \frac{y1  + y2}{2} )

U(8,9)

V(2,5)

m = [(8+2)/2 , (9+5)/2]

= [(10/2) , (14/2)]

= (5,7)

(Correct me if i am wrong)

5 0
3 years ago
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Lisa [10]

Answer:

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Step-by-step explanation:

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Answer:

The true statement is that the answer will be a fraction.

Step-by-step explanation:

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