Find the gradient of the function g(x,y)= xy² at the point (5, - 1). Then sketch the gradient together with the level curve that passes through the point.

First find the gradient vector at (5, - 1).

Vg(5,- 1) = [ ]i - [ ]j (Simplify your answers.)

Choose the graph that shows the level curve and the gradient vector at (5, - 1).

options of the sketched graphs are uploaded along this answers

Answer:

Vg(5,- 1) = [ 1 ]i - [ 10 ]j

Option A is the correct graph for this Level

Step-by-step explanation:

Given that;

g(x,y) = xy²

we have to find gradient of this function at ( 5, -1 )

Δg(x,y) = dg/dx + dg/dy

d(g)/dx = y² , dg/dy = 2gx

therefore

Δg(x,y) = [y²]i + [2yx]j

Δ( 5, -1) = [-1²]i + [2×-1 ×5]j

=Δ(5, -1) = 1i - 10j

[ 1 ]i - [ 10 ]j

Therefore Option A is the correct graph for this Level

3 0

3 years ago

For the function f(x)

WITCHER [35]

Given:

The function is

$f(x)=\dfrac{8\sqrt[3]{x}-8}{9}$

To find:

The inverse function $f^{-1}(x)$ .

Solution:

We have,

$f(x)=\dfrac{8\sqrt[3]{x}-8}{9}$

Step 1: Putting f(x)=y, we get

$y=\dfrac{8\sqrt[3]{x}-8}{9}$

Step 2: Interchange x and y.

$x=\dfrac{8\sqrt[3]{y}-8}{9}$

Step 3: Isolate variable y.

$9x=8\sqrt[3]{y}-8$

$9x+8=8\sqrt[3]{y}$

$\dfrac{9x+8}{8}=\sqrt[3]{y}$

Taking cube on both sides, we get

$\left(\dfrac{9x+8}{8}\right)^3=y$

$y=\left(\dfrac{9x+8}{8}\right)^3$

Step 4: Putting $y=f^{-1}(x)$ , we get

$f^{-1}(x)=\left(\dfrac{9x+8}{8}\right)^3$

Therefore, the correct option is C.

4 0

2 years ago