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frosja888 [35]
3 years ago
9

URGENT. PLEASE HELP WITH PROBLEM WITH EXPLANATION!!!

Mathematics
1 answer:
IRINA_888 [86]3 years ago
3 0

Answer:

D is the answer of the question

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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
6(-2.3x- 5) + (4x + 11)
Oxana [17]

x = -1.93877551

Idk but I tried

7 0
4 years ago
If 5 friends buy a package of 38 balloons, how should they divide them?
Nastasia [14]
All you have to do is divide 38 divided by 5 which = 7 r 3

I hope this helps. 

Have a awesome day. :)
7 0
3 years ago
Read 2 more answers
how to calculate cube root without using prime factorization? and how to calculate ³√60? or something else that has a decimal re
Marat540 [252]

The cube root of 60 is 3.87 approximately.

Step by step solution:

We can calculate the cube root by Halley's method:

The  formula is \sqrt[3]{a} = x ((x^{3}  + 2a)/(2x^{3}  + a))

where,

a = number whose cube root is being calculated

x = integer guess of its cube root.

Here a = 60,

Suppose x as 3

[∵ 3³ = 27 and 27 is the nearest perfect cube that is less than 60]

⇒ x = 3

Therefore,

∛60 = 3 (3³ + 2 × 60)/(2 × 3³ + 60)) = 3.87

⇒ ∛60 ≈ 3.87

Therefore, the cube root of 60 is 3.87 approximately.

Here ,  ∛60 is irrational because it cannot be expressed in the form of p/q where q ≠ 0.

Therefore, the value of the cube root of 60 is an irrational number.

Learn more about cube root :brainly.com/question/27863878

#SPJ1

4 0
2 years ago
Alyssa says that n = 6 is the solution of the equation 12n = 84. How can you check whether she is correct?
Vikki [24]

Answer:

No. The answer is 72

Step-by-step explanation:

As 12x6=72

4 0
3 years ago
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