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2 years ago

3+7+11+15+19+23+271.t1=s1=express t1 in terms of s1

Mathematics

1 answer:

icang [17]2 years ago

8 0

So it would be 3s + 7s + 11s + 15s + 19s + 23s + 27s. Hope this helps

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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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3 years ago

2x-3y=3 5x+2y=17 how do i solve this?

Alina [70]

2(2x-3y=3)..................4x-6y=6..........equation 1
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adding equation 1 and equation 2;
19x=57
x=3

2x-3y=3
(2*3)-3y=3
6-3=3y
y=1

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3 years ago

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Answer:

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Answer:

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3+7+11+15+19+23+271.t1=s1=express t1 in terms of s1​

3+7+11+15+19+23+271.t1=s1=express t1 in terms of s1