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3 years ago

What point is on the graph of f(x)=3x4

Mathematics

1 answer:

Artyom0805 [142]3 years ago

5 0

Answer:

There are an infinite number of points on this graph. Two of them would be (1, 3) and (0, 0).

Step-by-step explanations:

In order to find any points on the graph, simply put the x value in to get the y value.

f(x) = 3x^4

f(x) = 3(1)^4

f(x) = 3(1)

f(x) = 3

(1, 3)

And then for the second point.

f(x) = 3x^4

f(x) = 3(0)^4

f(x) = 3(0)

f(x) = 0

(0, 0)

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Use the Chain Rule to find the indicated partial derivatives. z = x^4 + xy^3, x = uv^4 + w^3, y = u + ve^w Find : ∂z/∂u , ∂z/∂v

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I'll use subscript notation for brevity, i.e. $\frac{\partial f}{\partial x}=f_x$ .

By the chain rule,

$z_u=z_xx_u+z_yy_u$

$z_v=z_xx_v+z_yy_v$

$z_w=z_xx_w+z_yy_w$

We have

$z=x^4+xy^3\implies\begin{cases}z_x=4x^3+y^3\\z_y=3xy^2\end{cases}$

and

$\begin{cases}x=uv^4+w^3\\y=u+ve^w\end{cases}\implies\begin{cases}x_u=v^4\\x_v=4uv^3\\x_w=3w^2\\y_u=1\\y_v=e^w\\y_w=ve^w\end{cases}$

When $u=1,v=1,w=0$ , we have

$\begin{cases}x(1,1,0)=1\\y(1,1,0)=2\end{cases}\implies\begin{cases}z_x(1,2)=12\\z_y(1,2)=12\end{cases}$

and the partial derivatives take on values of

$\begin{cases}x_u(1,1,0)=1\\x_v(1,1,0)=4\\x_w(1,1,0)=0\\y_u(1,1,0)=1\\y_v(1,1,0)=1\\y_w(1,1,0)=1\end{cases}$

So we end up with

$\boxed{\begin{cases}z_u(1,1,0)=24\\z_v(1,1,0)=60\\z_w(1,1,0)=12\end{cases}}$

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3 years ago