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Mathematics

1 answer:

kipiarov [429]3 years ago

6 0

I had this question on my test>>>>>>>>> v=5y+22

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

7 0

3 years ago

Camden invested $260 in an account paying an interest rate of 4 1/8% compounded annually. Evan invested $260 in an account payin

Archy [21]

Answer:

Step-by-step explanation

got it wrong so many times so delta math gave me the answer

5 0

3 years ago

Solve: (-5/6) × (9/20) + (-5/6) × 7/25 = ?

jonny [76]

$\bf \underline{★ How \:to\: do -} \\$

Here, we are given with four fractions to multiply two of them and to add two of them. If we add them directly by taking the LCM and adding them is not a similar way. We can clearly observe that in those four fractions, we have two fractions as common i.e, we have two fractions as same. If we have two fractions or numbers as same, we can solve the sum by an other concept called as distributive property. In this property, we multiply the common fraction with the sum of other two fractions. This concept can also be done with fractions as well as integers. So, let's solve!!

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