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3 years ago

Plz help me!!!!!!!!!!

Mathematics

2 answers:

Vlad1618 [11]3 years ago

4 0

Answer: $\bold{\dfrac{5\pm \sqrt{57}}{4}}$

Step-by-step explanation: $2a-5=\dfrac{4}{a}\\\\\text{Multiply by the LCD (a) to clear the denominator:}\\\\\\2a(a)-5(a)=\dfrac{4}{a}(a)\\\\2a^2-5a=4\\\\2a^2-5a-4=0\quad \rightarrow \quad a=2,\ b=-5,\ c=-4\\\\\\\text{Quadratic formula is: }x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4(2)(-4)}}{2(2)}\\\\\\.\ =\dfrac{5\pm \sqrt{25+32}}{4}\\\\\\.\ =\dfrac{5\pm \sqrt{57}}{4}$

avanturin [10]3 years ago

3 0

Answer:

a = ( 5 +/- sqrt(57) )/2

Step-by-step explanation:

Uh...well just multiply by a and move the 4, setting the equation to 0.

a, b, and c are 2, -5, -4

Plug that into the QuadF.

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Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify

irina [24]

Answer:

x-coordinates of relative extrema = $\frac{-1}{2}$

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

$f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}$

Differentiate with respect to x

$f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}$

$f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}$

Differentiate f'(x) with respect to x

$f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1$

At x = $\frac{-1}{2}$ ,

$f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0$

We know that if $f''(a)>0$ then x = a is a point of minima.

So, $x=\frac{-1}{2}$ is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

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4 years ago

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3 years ago

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Snowcat [4.5K]

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2 years ago

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