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3 years ago

If f (x) = x-2 and g (x,y) = y2 + x, then g (3, f(4)) is?

Mathematics

2 answers:

alex41 [277]3 years ago

4 0

$f(4)=4-2=2\\\\
g(3,f(4))=g(3,2)=2^2+3=4+3=7$

aleksandr82 [10.1K]3 years ago

4 0

F(x) = x -2

f(4) = 4 - 2 = 2

g(x , y) = y² + x

g(3, f(4))

g(3 , 2) = 2² + 3 = 4 + 3 = 7

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Step-by-step explanation: Hope this help :D

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Use implicit differentiation to find the slope of the tangent line at the given point:

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Answer:

$\frac{dy}{dx}=0$

Step-by-step explanation:

So we have the equation:

$(x^2+y^2)^2=4x^2y$

And we want to find the slope of the tangent line at the point (1,1).

So, let's implicitly differentiate. Take the derivative of both sides:

$\frac{d}{dx}[(x^2+y^2)^2]=\frac{d}{dx}[4x^2y]$

Let's do each side individually.

Left:

We can use the chain rule:

$(u(v(x))'=u'(v(x))\cdot v'(x)$

Let's let v(x) be x²+y². So, u(x) is x². Thus, the u'(x) is 2x. Therefore:

$\frac{d}{dx}[(x^2+y^2)^2]=2(x^2+y^2)(\frac{d}{dx}[x^2+y^2])$

We can differentiate x like normal. However, for y, we must differentiate implicitly. pretend y is y(x). This gives us:

$\frac{d}{dx}[(x^2+y^2)^2]=2(x^2+y^2)(\frac{d}{dx}[x^2]+\frac{d}{dx}[y^2(x)])$

Differentiate:

$\frac{d}{dx}[(x^2+y^2)^2]=2(x^2+y^2)(2x+2y\frac{dy}{dx})$

Therefore, our left side is:

$2(x^2+y^2)(2x+2y\frac{dy}{dx})$

Right:

We have:

$\frac{d}{dx}[4x^2y]$

Let's move the 4 outside:

$=4\frac{d}{dx}[x^2y]$

Use the product rule:

$=4(\frac{d}{dx}[x^2]y+x^2\frac{d}{dx}[y])$

Differentiate:

$=4(2xy+x^2\frac{dy}{dx})$

Therefore, our entire equation is:

$2(x^2+y^2)(2x+2y\frac{dy}{dx})=4(2xy+x^2\frac{dy}{dx})$

So, to find the derivative at (1,1), substitute 1 for x and 1 for y.

$2((1)^2+(1)^2)(2(1)+2(1)\frac{dy}{dx})=4(2(1)(1)+(1)^2\frac{dy}{dx})$

Evaluate.

$2((1)+(1))(2+2\frac{dy}{dx})=4(2+\frac{dy}{dx})$

Simplify. Also, let's distribute the right:

$2(2)(2+2\frac{dy}{dx})=8+4\frac{dy}{dx}$

Multiply.

$4(2+2\frac{dy}{dx})=8+4\frac{dy}{dx}$

Distribute the left:

$8+8\frac{dy}{dx}=8+4\frac{dy}{dx}$

Subtract 8 from both sides:

$8\frac{dy}{dx}=4\frac{dy}{dx}$

Subtract 4(dy/dx) from both sides:

$4\frac{dy}{dx}=0$

Divide both sides by 4:

$\frac{dy}{dx}=0$

Therefore, the slope at the point (1,1) is 0.

And we're done!

We can verify this using the graph. The slope of the line tangent to the point (1,1) seems like it would be horizontal, giving us a slope of 0.

Edit: Typo

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An arithmetic sequence is one where each term is a constant difference, called the common difference, from the preceding term. The arithmetic sequence can always be expressed as:

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