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mezya [45]
3 years ago
5

EXPONENTIAL FUNCTIONS HELP Write the function for each graph described below. the graph of f(x) = 2x reflected across the x-axis

. The graph of f(x)= 1/3x translated up 5 units. The graph of f(x) = 3x left 2 units, and down 3. The graph of f(x) = 1/2x translated down 2 units. The graph of f(x) = 4x stretched horizontally by a factor of 3. The graph of f(x) = 2x up 4 units, right 3.
Mathematics
1 answer:
raketka [301]3 years ago
7 0

Answer:

  • -2^x
  • (1/3)^x +5
  • 3^(x +2) -3
  • (1/2)^x -2
  • 4^(x/3)
  • 2^(x -3) +4

Step-by-step explanation:

In general, the transformation ...

  g(x) = f(x -h) +k

translates f(x) right h units and up k units.

The transformation ...

  g(x) = f(x/a)

stretches the graph horizontally by a factor of "a".

The transformation ...

  g(x) = -f(x)

causes the graph to be reflected over the x-axis.

___

Applying the above, we have ...

f(x) = 2^x reflected over x is g(x) = -2^x

f(x) = (1/3)^x translated up 5 is g(x) = (1/3)^x +5

f(x) = 3^x translated by (-2, -3) is g(x) = 3^(x +2) -3

f(x) = (1/2)^x translated down 2 is g(x) = (1/2)^x -2

f(x) = 4^x stretched horizontally by a factor of 3 is g(x) = 4^(x/3)

f(x) = 2^x translated by (3, 4) is g(x) = 2^(x -3) +4

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Answer: Absolute minimum: f(-1) = -2\sqrt{6}

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Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = t\sqrt{25-t^{2}}

f'(t) = 1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)

f'(t) = \sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }

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25-2t^{2} = 0

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\sqrt{25-t^{2}} ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = -\sqrt{2.5} and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = t\sqrt{25-t^{2}}

f(-1) = -1\sqrt{25-(-1)^{2}}

f(-1) = -\sqrt{24}

f(-1) = -2\sqrt{6}

f(\sqrt{12.5}) = \sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}

f(\sqrt{12.5}) = 12.5

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f(5) = 0

Therefore, absolute maximum is f(\sqrt{12.5}) = 12.5 and absolute minimum is

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