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3 years ago

Find the points on the curve y = 2x3 + 3x2 − 12x + 6 where the tangent line is horizontal.

1 answer:

3 0

N.b., this is copied from an answer I wrote for someone else.

Thinking about the graph of y, the rate of change is zero whenever there is an extremum.

First, differentiate y.

$y'=6x^2+6x-12$

Next, find the zeros of y' by factoring.

$6(x^2+x-2)=0 \\ x = -2, x = 1$

Now, we substitute these x-values into the original equation to find the coordinates.

$(-2,21), (1,-6)$

You might be interested in

The height of a cone is twice the radius of its base.

fredd [130]

Answer: 2/3πx³

Step-by-step explanation:

Let the radius of the cone be represented by x.

Since the height of the cone is twice the radius of its base, the height will be: = 2x

Volume of a cone = 1/3πr²h

where,

r = x

h = 2x

Volume of a cone = 1/3πr²h

= 1/3 × π × x² × 2x

= 1/3 × π × 2x³

= 2/3πx³

Therefore, the correct answer is 2/3πx³.

3 0

3 years ago

The sum of the 6th and seventh terms in a sequence of 12 numbers with a common difference is 16. What is the Sum of the number s

patriot [66]

Answer:

Step-by-step explanation:

commom difference = d

a₆ + a₇ = 16

a₅ + a₈ = (a₆ - d) + (a₇ + d) = a₆ + a₇ = 16

a₄ + a₉ = (a₆ - 2d) + (a₇ + 2d) = a₆ + a₇ = 16

Similarly,

a₃ + a₁₀ = 16, a₂ + a₁₁ = 16, a₁ + a₁₂ = 16

a₁ + a₂ + ... + a₁₁ + a₁₂ = 6 x 16 = 96

6 0

1 year ago

The scale model of a rectangular garden is 1.5 ft by 4 ft. The scale model is enlarged by a scale factor of 7 to create the actu

KengaRu [80]

I will say 42, I am sorry if I am incorrect.

4 0

3 years ago

Read 2 more answers

Ignoring leap days, the days of the year can be numbered 1 to 365. Assume that birthdays are equally likely to fall on any day o

faust18 [17]

Answer:

Follows are the solution to the given question:

Step-by-step explanation:

They can count the days of the year 1 to 365. The random project consists of drawing a sample of n objects from D where elements are n people's birth in a group but instead, D = {1,....365}. And then there's the issue.

$S=365^n$

This because the list of future birthdays of n people was its test point; therefore m points will be in the sequence so each point contains 365 distinct outcomes. The probability function P for \Omega is that any event is likely to happen in 365 days.

$P(x)=\frac{1}{365^{n}}$

if x is between 1 and 365 as well as the occurrence is just all similarly possible

In point i:

That somebody mentions their birthday throughout the party

Guess I was born on day b. Therefore the consequence of "x is in A" is "b is now in the series of x," which is to say, b = bk for some amount k approximately 1 and n.

In point ii:

Any 2 persons share the same birthday at this party". A result x is in B" means which "two of entries in x are same." This means that perhaps the outcome x is in B if or only if bj = bk is in B of two numbers j, and k of 1, of two. , no, n.

In point iii:

Many three students share the same birthday with both the party. The consequence is x at the level of C only when bj = bk = bl at three (different) indices, j, k, l, 1. , no, n.

6 0

2 years ago

Which is equivalent to 64 Superscript one-fourth?

Schach [20]

Answer:

2 RootIndex 4 StartRoot 4 EndRoot

Step-by-step explanation:

we have

$64^{\frac{1}{4}}$