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2 years ago

Does the point (3 , 4) lie on the liney = 2x -1?

Mathematics

1 answer:

belka [17]2 years ago

5 0

Answer:

Step-by-step explanation:

y = 2x -1

Substitute the point into the equation

4 = 2(3) -1

4 = 6-1

4 =5

This is not a true statement so the point is not on the line

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alexira [117]

Answer:

The inner function is $h(x)=4x^2 + 8$ and the outer function is $g(x)=3x^5$ .

The derivative of the function is $\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=120x\left(4x^2+8\right)^4$ .

Step-by-step explanation:

A composite function can be written as $g(h(x))$ , where $h$ and $g$ are basic functions.

For the function $f(x)=3(4x^2+8)^5$ .

The inner function is the part we evaluate first. Frequently, we can identify the correct expression because it will appear within a grouping symbol one or more times in our composed function.

Here, we have $4x^2+8$ inside parentheses. So $h(x)=4x^2 + 8$ is the inner function and the outer function is $g(x)=3x^5$ .

The chain rule says:

$\frac{d}{dx}[f(g(x))]=f'(g(x))g'(x)$

It tells us how to differentiate composite functions.

The function $f(x)=3(4x^2+8)^5$ is the composition, $g(h(x))$ , of

outside function: $g(x)=3x^5$

inside function: $h(x)=4x^2 + 8$

The derivative of this is computed as

$\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=3\frac{d}{dx}\left(\left(4x^2+8\right)^5\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\f=u^5,\:\:u=\left(4x^2+8\right)\\\\3\frac{d}{du}\left(u^5\right)\frac{d}{dx}\left(4x^2+8\right)\\\\3\cdot \:5\left(4x^2+8\right)^4\cdot \:8x\\\\120x\left(4x^2+8\right)^4$

The derivative of the function is $\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=120x\left(4x^2+8\right)^4$ .

3 0

3 years ago

Does the point (3 , 4) lie on the liney = 2x -1?​

Does the point (3 , 4) lie on the liney = 2x -1?