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tiny-mole [99]
3 years ago
11

Y=x+ - 4x3 Find the two stationary points on the graph of y=x4 - 4x3

Mathematics
1 answer:
mafiozo [28]3 years ago
5 0

Answer: (0,0), (3,-27)

Step-by-step explanation:

Given

Curve is y=x^4-4x^3

The stationary point on a differentiable function is the points where the differentiation of the function is zero i.e. slope is zero at that point.

Differentiate the curve f(x)=x^4-4x^3

\Rightarrow f'(x)=4x^3-12x^2

Equate it to zero

\Rightarrow 4x^3-12x^2=0\\\Rightarrow 4x^2(x-3)=0\\\Rightarrow x=0,0,3

Put x=0,3 in the function f(x)=x^4-4x^3

\Rightarrow f(0)=0\\\Rightarrow f(3)=3^4-4(3)^3\\\Rightarrow f(3)=81-108\\\Rightarrow f(3)=-27

Therefore, the stationary points are (0,0), (3,-27)

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2x^2+5x+3/x^2-3x-4 divided by 4x^2+2x-6/x^2-8x+16
Masja [62]

Answer: \frac{x-4}{2x-2}

Step-by-step explanation:

\frac{2x^2+5x+3}{x^2-3x-4} divided by \frac{4x^2+2x-6}{x^2-8x+16} is the same thing as multiplying \frac{2x^2+5x+3}{x^2-3x-4} by \frac{x^2-8x+16}{4x^2+2x-6}.

The equation we get is:

\frac{2x^2+5x+3}{x^2-3x-4}*\frac{x^2-8x+16}{4x^2+2x-6}

With this equation, we can factor each equation:

\frac{(x+1)(2x+3)}{(x+1)(x-4)} *\frac{(x-4)(x-4)}{2(x-1)(2x+3)}

We can cancel out like terms since they would be dividing each other:

\frac{x-4}{2x-2}

8 0
3 years ago
Solve the system of equations.<br> 2y+7x=−5<br> 5y−7x=12<br> ​
3241004551 [841]

We can solve this by substitution method.

Look at the second equation. If we rearrange to find 7x, we can substitute in the value into the first equation.

5y-7x=12

5y-7x-12=0

5y-12=7x

Therefore, 7x=5y-12

Now replace the 7x in the first equation with 5y - 12:

2y+7x=-5 (substitute in 7x = 5y - 12)

2y+(5y-12)=-5

7y-12=-5

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y=1

Now that we know y, we can find x by substituting in y = 1 into any equation we want. I will use the equation: 7x = 5y - 12

7x=5y-12 (substitute in y = 1)

7x=5(1) -12

5x=5-12

7x=-7

x=-1

__________________________________________________________

<u>Answer:</u>

<u></u>y=1\\x=-1<u></u>

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Step-by-step explanation:

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