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Llana [10]
3 years ago
11

Solve the following system. y = (1/2)x^2 + 2x - 1 and 3x - y = 1 The solutions are (____)and (______ ) (remember to include the

commas)
Mathematics
2 answers:
lubasha [3.4K]3 years ago
7 0

Answer:

x=0, 2. y=-1, 5.

Step-by-step explanation:

y=(1/2)x^2+2x-1

3x-y=1

---------------------

3x-y=1

y=3x-1

-----------

(1/2)x^2+2x-1=3x-1

(1/2)x^2+2x-3x-1-(-1)=0

(1/2)x^2-x-1+1=0

(1/2)x^2-x=0

factor out the x

x[(1/2)x-1]=0

zero property,

x=0, (1/2)x-1=0

-------------------

(1/2)x-1=0

1/2x=0+1

1/2x=1

x=1/(1/2)=(1/1)(2/1)=2/1=2

-----------------------------------

x=0, x=2

-------------

y=(1/2)(0)^2+2(0)-1=0+0-1=-1

-------------

y=3(0)-1=0-1=-1

-----------------------

y=(1/2)(2)^2+2(2)-1=(1/2)(4)+4-1=4/2+4-1=2+4-1=6-1=5

----------------------

y=3(2)-1=6-1=5

ASHA 777 [7]3 years ago
7 0

Answer:

The solutions are (0, -1) and (2, 5)

Step-by-step explanation:

y = (1/2)x^2 + 2x - 1 ------ eqn(I)

3x - y = 1

y = 3x - 1 ------------- eqn(II)

Equate eqn(I) & (II)

(1/2)x^2 + 2x - 1 = 3x - 1

Multiply each term by 2

x^2 + 4x - 2 = 6x - 2

x^2 + 4x - 6x = -2 + 2

x^2 - 2x = 0

x(x - 2) = 0

x = 0, 2

Substitute the values of x in eqn(II)

y = 3x - 1

When x = 0

y = 3(0) - 1 = 0 - 1 = -1

y = 3x - 1

When x = 2

y = 3(2) - 1 = 6 - 1 = 5

The solutions are (0, -1) and (2, 5)

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<u>Given</u>:

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<u>First derivative</u>

\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If  $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}

<u />

<u />\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If  $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}

<u />

\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1}  \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}

<u>Second derivative</u>

<u />

\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If  $y=uv$  then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}

\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}

\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}

\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m

\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m

\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}

              = \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)

<u>Proof</u>

  (x^2+1)y_2+xy_1-m^2y

= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m

= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m

= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]

= \left(x+\sqrt{1+x^2}\right)^m\left[0]

= 0

8 0
1 year ago
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