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erastova [34]
3 years ago
7

G(x) = 3x - 3 n(x) = 2x - 3 Find (g. h)(x)

Mathematics
1 answer:
Aleksandr [31]3 years ago
6 0

<em>Greetings From Brasil...</em>

Be

G(X) = 3X - 3

H(X) = 2X - 3

(GoH)(X) = G(H(X))

So, just replace the X of G(X) by the function H(X)

G(X) = 3X - 3

G(H(X)) = 3 · H(X) - 3

G(H(X)) = 3 · (2X - 3) - 3

G(H(X)) = (6X - 9) - 3

<h2>G(H(X)) = 6X - 12</h2>

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The first 2 and the last one

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2 years ago
For which equations is 8 a solution? Select the four correct answer
maxonik [38]

Answer:

I <em>believe</em> the answers are= 2, 3, 6, 8

Step-by-step explanation:

For #2, you would subtract 2 from both sides. 10-2=8. x=8.

For #3, you would add 4 to both sides of the equation, making the final answer: x=8

For #6, you would divide both sides by 3. 24 divided by three is 8.

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Hope this helps!  

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3 years ago
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Find the local max and min values of f(x)=x^2/x-1 using both first and second derivative tests
uranmaximum [27]
Hmm, the 2nd derivitve is good for finding concavity

let's find the max and min points
that is where the first derivitive is equal to 0
remember the difference quotient

so
f'(x)=(x^2-2x)/(x^2-2x+1)
find where it equals 0
set numerator equal to 0
0=x^2-2x
0=x(x-2)
0=x
0=x-2
2=x

so at 0 and 2 are the min and max
find if the signs go from negative to positive (min) or from positive to negative (max) at those points

f'(-1)>0
f'(1.5)<0
f'(3)>0

so at x=0, the sign go from positive to negative (local maximum)
at x=2, the sign go from negative to positive (local minimum)


we can take the 2nd derivitive to see the inflection points
f''(x)=2/((x-1)^3)
where does it equal 0?
it doesn't
so no inflection point
but, we can test it at x=0 and x=2
at x=0, we get f''(0)<0 so it is concave down. that means that x=0 being a max makes sense
at x=2, we get f''(2)>0 so it is concave up. that means that x=2 being a max make sense




local max is at x=0 (the point (0,0))
local min is at x=2 (the point (2,4))
6 0
3 years ago
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