F(x)+x^2-3x-2x is shifted 4 units left. The result is g(x). What is g(x)

Mathematics

1 answer:

otez555 [7]3 years ago

3 0

Answer:

g(x) = x² + 5x + 2

Step-by-step explanation:

If a function is f(x) = x²

And this function is shifted 4 units left then the rule to be followed is,

g(x) = f(x + 4)

Following the same rule,

For a function f(x) = x² - 3x - 2 when shifted 4 units to the left,

g(x) = f(x + 4)

g(x) = (x + 4)² - 3(x + 4) - 2

= x² + 8x + 16 - 3x - 12 - 2

= x² + 5x + 2

Therefore, g(x) = x² + 5x + 2 will be the transformed function.

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The full question is shown below

a_sh-v [17]

Answer:

n = 3

Step-by-step explanation:

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8 0

3 years ago

Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth

taurus [48]

Answer:

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

$y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\$

It is a homogeneous solution:

$y_h(t)=c_1e^{-2i t}+c_2e^{2i t}$

Now, we finding a particular solution.

$y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\$

we get

$y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$