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3 years ago

need help. solve for y. try factoring first if not possible or difficult use quadratic formula. y²-6y+6=0

Mathematics

2 answers:

shutvik [7]3 years ago

6 0

$y^2-6y+6=0\\
y^2-6y+9-3=0\\
(y-3)^2=3\\
|y-3|=\sqrt3\\
y-3=\sqrt3 \vee y-3=-\sqrt3\\
y=3+\sqrt3 \vee y=3-\sqrt3$

lbvjy [14]3 years ago

3 0

$y^2-6y+6=0\\ \\a=1 , \ b=-6, \ c=6 \\ \\\Delta =b^2-4ac = (-6)^2 -4\cdot1\cdot6 = 36-24=12\\ \\\sqrt{\Delta }= \sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3} \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{6-2\sqrt{3}}{2 }=\frac{2( 3- \sqrt{3})}{2}= 3- \sqrt{3}$

$x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{6+2\sqrt{3}}{2 }=\frac{2( 3+ \sqrt{3})}{2}= 3+ \sqrt{3}$

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4 years ago

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Part A:

Given the function $f(x)=x+2\cos(x)$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow1-2\sin{x}=0 \\ \\ \Rightarrow2\sin{x}=1 \\ \\ \Rightarrow\sin{x}= \frac{1}{2} \\ \\ \Rightarrow x=\sin^{-1}{\frac{1}{2}}= \frac{\pi}{6}$

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$f''(x)=-2cosx \\ \\ \Rightarrow f''\left( \frac{\pi}{6} \right)=-2\cos{\left( \frac{\pi}{6} \right)}=-1.732$

Since the second derivative gives a negative number, the given function has a maximum point at $x=\frac{\pi}{6}$ .

And the maximum point is given by:

$f\left( \frac{\pi}{6} \right)=\frac{\pi}{6}+2\cos\left( \frac{\pi}{6} \right) \\ \\ =0.5236+2(0.8660)=0.5236+1.732 \\ \\ =\bold{2.256}$

i.e. $\left(\frac{\pi}{6},\ 2.256\right)$

Part B:

Given the function $f(x)=e^{-x}-e^{-2x}$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow-e^{-x}+2e^{-2x}=0 \\ \\ \Rightarrow2e^{-2x}=e^{-x} \\ \\ \Rightarrow2e^{-x}=1 \\ \\ \Rightarrow e^{-x}=\frac{1}{2} \\ \\ \Rightarrow-x=\ln \frac{1}{2}=-0.6931 \\ \\ \Rightarrow x=0.6931$

Using the second derivative test,

$f''(x)=e^{-x}-4e^{-2x} \\ \\ \Rightarrow f''(0.6931)=e^{-0.6931}-4e^{-2(0.6931)} \\ \\ =0.5-4e^{-1.386}=0.5-4(0.25)=0.5-1 \\ \\ =-0.5$

Since the second derivative gives a negative number, the given function has a maximum point at $x=0.6931$ .

And the maximum point is given by:

$f(0.6931)=e^{-0.6931}-e^{-2(0.6931)} \\ \\ =0.5-e^{-1.386}=0.5-0.25=\bold{0.25}$

i.e. (0.693, 0.25)

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3 years ago