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anygoal [31]
3 years ago
7

qrt{x - 1} (x - 6) {}^{2} " alt=" \sqrt{x - 1} (x - 6) {}^{2} " align="absmiddle" class="latex-formula">
inflection points?​
Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
7 0

Answer:

1 and 6

Step-by-step explanation:

it is just like the roots of a polynomial, since \sqrt{x-1}(x-6)^2=0

we can separate the two pieces of the problem

\sqrt{x-1} =0

square both sides to get

x-1=0.   x=1

(x-6)^2=0

square root both sides and get

x-6=0

x=6

so the two inflection points of the equation are 6 and 1

Hope that helps :)

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Answer:

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Step-by-step explanation:

f$ \circ  $ g(x) = f(g(x))

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and       g(x) = x - 2

Now f(g(x)) = f(x - 2) = 2(x - 2)²

We know that (a - b)² = a² - b² + 2ab

Using this we expand f(g(x)). We get:

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Similarly,        g(f(x)) = g(2x²) = 2x² - 2

Now, f(g(-2))    = 2[(-2)² - 4(-2) + 4]     = 2(16)                           = 32.

Also, g(f(-2))   = 2[(-2)² - 2]                = 2(2)                             = 4.

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g(f(3.5))  = 2{(3.5)² -2} = 2{12.25 - 2}                     = 2(10.25)   = 20.5.

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g(f(0))     = 2{0 - 2}                         = 2(-2)                                 = -4.

Arranging them in ascending order, we get:

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