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Oliga [24]
4 years ago
13

The point (6, 6) is on the graph y = f(x) . Find the corresponding coordinates of this point on the graph y = 4f[1/3x +9] -7

Mathematics
1 answer:
miss Akunina [59]4 years ago
8 0

Answer:

Correct answer is:

<em>a. (-9,17)</em>

Step-by-step explanation:

We are given that a point (6, 6) lies on the graph of y =f(x).

Putting the values from the given point:

6 =f(6)

That means we are given that f(6) =6 ..... (1)

And we have to find the corresponding coordinates of this point on the graph of y = 4f[\frac{1}3x +9] -7

From equation (1), we know the value of f(6).

so, let us convert f[\frac{1}3x +9] to a form such that it becomes equal to f(6)

\Rightarrow \dfrac{1}{3}x +9 =6\\\Rightarrow \dfrac{1}{3}x=-3\\\Rightarrow x = -9

So, let us put x = -9 in the given function:

4f[\frac{1}3\times (-9) +9] -7\\\Rightarrow 4f[-3 +9] -7\\\Rightarrow 4f(6) -7

Now, using equation <em>(1)</em>, putting f(6)=6

\Rightarrow 4\times 6 -7\\\Rightarrow 24-7 \\\Rightarrow y = 17

Therefore, the point the corresponding point is:

<em>a. (-9,17)</em>

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artcher [175]

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f(x, y) = Sin(x*y)

We want the second order taylor expansion around x = 0, y = 0.

This will be:

f(x,y) = f(0,0) + \frac{df(0,0)}{dx} x + \frac{df(0,0)}{dy} y + \frac{1}{2} \frac{d^2f(0,0)}{dx^2} x^2 +\frac{1}{2} \frac{d^2f(0,0)}{dy^2}y^2  + \frac{d^2f(0,0)}{dydx} x*y

So let's find all the terms:

Remember that:

\frac{dsin(ax)}{dx}  = a*cos(ax)

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f(0,0) = sin(0*0) = 1.

\frac{df(0,0)}{dx}*x = y*cos(0*0)*x = x*y

\frac{df(0,0)}{dy} *y = x*cos(00)*y = x*y

\frac{1}{2} \frac{d^2f(0,0)}{dx^2}*x^2 =  -\frac{1}{2}  *y^2*sin(0*0)*x^2 = 0

\frac{1}{2} \frac{d^2f(0,0)}{dy^2}*y^2 =  -\frac{1}{2}  *x^2*sin(0*0)*y^2 = 0

\frac{d^2f(0,0)}{dxdy} x*y = (cos(0*0) -x*y*sin(0*0))*x*y = x*y

Then we have that the taylor expansion of second order around x = 0 and y = 0 is:

sin(x,y) = x*y + x*y + x*y = 3*x*y

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