Exponential Growth Functions

Mathematics

1 answer:

Ainat [17]1 year ago

7 0

F (x) = 5(4) ^x Is the right answer

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4. The cube root parent function is vertically stretched by a factor of 3, reflected in the x-axis,

Mademuasel [1]

Using translation concepts, it is found that the equation that represents the new function is:

$y = -3\sqrt[3]{x} - 8$

The cube root parent function is given by:

$y = \sqrt[3]{x}$

<h3>Translation:</h3>

The function goes through these following translations:

Vertically stretched by a factor of 3, that is, it is multiplied by 3, hence $y = 3\sqrt[3]{x}$ .

Reflected over the x-axis, that is, it is multiplied by -1, hence $y = -3\sqrt[3]{x}$ .

Translated 8 units down, that is, 8 is subtracted from the function, hence $y = -3\sqrt[3]{x} - 8$ .

Then, the equation is:

$y = -3\sqrt[3]{x} - 8$

To learn more about translation concepts, you can take a look at brainly.com/question/13671886

3 0

2 years ago

Let R be the region bounded by the following curves. Let S be the solid generated when R is revolved about the given axis. Ifpo

steposvetlana [31]

The two curves $y=x$ and $y=x^{1/15}$ intersect at $x=0$ and $x=1$ , with $x^{1/15}\ge x$ over $0\le x\le1$ .

Washer method

$\displaystyle\pi\int_0^1\left(\left(x^{1/15}\right)^2-x^2\right)\,\mathrm dx=\pi\int_0^1\left(x^{2/15}-x^2\right)\,\mathrm dx=\pi\left(\frac{15}{17}-\frac13\right)=\boxed{\frac{28\pi}{51}}$

Shell method

We have $y=x^{1/15}\implies x=y^{15}$ . The curves $x=y$ and $x=y^{15}$ intersect at $y=0$ and $y=1$ , with $y\ge y^{15}$ over $0\le y\le1$ .

$\displaystyle2\pi\int_0^1y\left(y-y^{15}\right)\,\mathrm dy=2\pi\int_0^1\left(y^2-y^{16}\right)\,\mathrm dy=2\pi\left(\frac13-\frac1{17}\right)=\boxed{\frac{28\pi}{51}}$