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nasty-shy [4]
3 years ago
5

Can someone double check my work #8

Mathematics
1 answer:
solong [7]3 years ago
6 0
Looks fine to me! I hope this helped
You might be interested in
consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where
boyakko [2]

Answer:

a)X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

b)Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

Step-by-step explanation:

From the question we are told that

The Function

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

Generally the differentiation of function f(x) is mathematically solved as

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

f(x)=\frac{x^3+x^2+5x+1}{x^2}

Therefore

f'(x)=\frac{x^2+10x+3}{x^4}

Generally critical point is given as

f'(x)=0

\frac{x^2+10x+3}{x^4}=0

x=-5 \pm\sqrt{22}

Generally the maximum and minimum x value for critical point is mathematically solved as

f'(-5 \pm\sqrt{22})

Where

Maximum value of x

f'(-5 +\sqrt{22})

Minimum value of x

f'(-5 +\sqrt{22})

Therefore interval of increase is mathematically given by

f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})

f(x)

Therefore interval of decrease is mathematically given by

(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)

Generally the second differentiation of function f(x) is mathematically solved as

f''(x)=\frac{2(x^2+15x+6)}{x^5}

Generally the point of inflection is mathematically solved as

f''(x)=0

x^2+15x+6=0

Therefore inflection points is given as

x=\frac{1}{2} (-15 \pm \sqrt{201}

f''(x)>0,\frac{1}{2}(-15-\sqrt{201})

a)Generally the concave upward interval X is mathematically given as

X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

f''(x)

b)Generally the concave downward interval Y is mathematically given as

Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

5 0
3 years ago
HELP PLSSSSS ON KHAN ACADEMY
torisob [31]

\boxed{\large{\bold{\blue{ANSWER~:) }}}}

See this attachment

4 0
3 years ago
9) Let<br> X-2, x, x+3 <br> be the first 3 terms of a geometric sequence. FInd the value of x..
Reil [10]
Hello,

Let's r the ration

We suppose x ≠0

x=r*(x-2)==>r=x/(x-2)

x+3=r*x==> (x+3)/x=x/(x-2)==>(x+3)(x-2)=x²
==>x²+3x-2x-6=x² ==>x-6=0==>x=6

x-2=6-4=4
x=6
x+3=6+3=9
6=4*3/2
9=6*3/2

Ration 3/2
First term: 4

7 0
3 years ago
What are the values of h and x in the right triangle? Explain.
alexandr402 [8]

Answer:

h = 60 ; x = 25

Step-by-step explanation:

You are going to want to use Pythagorean Theorem to solve these.

Pythagorean Theorem: a² + b² = c²

h² + 144² = 156²

h² + 20736 = 24336

h² = 3600

h = 60

x² + 60² = 65²

x² + 3600 = 4225

x² = 625

x = 25

3 0
3 years ago
Can you help
snow_tiger [21]
Positive if I looked at graph correctly
8 0
3 years ago
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