3 years ago

Find all the values of x where the tangent line is horizontal. 2x^3+36x^2+192x+12

1 answer:

3 0

At x = -4 and -8
Take the derivative of the given equation to find rate of change of the graph.
f'(x) = 6x^2 + 72x + 192
At a horizontal tangent, f'(x) = 0, so set the equation equal to 0 and solve. Eventually you get x = -4 and -8

You might be interested in

How can i solve this

lana66690 [7]

X+3-2x+5=10
+2x +2x
3x+ 3 + 5 = 10
-3 -3
3x + 5 = 7
-5 -5
3x=2
x=1.5
I hope all is well, and I helped! (: Good luck, rockstar!

8 0

3 years ago

How many terms are in this expression

Natali5045456 [20]

In number 2 there are 4 which are 22,x,3 and y. In number 4 there are 4 which are a,b,c and d.

3 0

3 years ago

Sin(cos^-1 x)<br><br> how to solve?

tatiyna

Answer:

$sin(cos^{-1}x)=\sqrt{1-x^2}$

Step-by-step explanation:

let us assume

$cos^{-1} x=\theta\\x=cos\theta\\$

now

$sin(cos^{-1}x)$ can be written as $sin\theta$

we know $sin^2\theta+cos^2\theta=1$

so $sin^2\theta=1-cos^2\theta$

$sin\theta=\sqrt{1-cos^2\theta}$

now from above equation we can say that

$sin\theta=\sqrt{1-x^2}$

$sin(cos^{-1}x)=\sqrt{1-x^2}$

7 0

3 years ago

I’ll mark brainliest please help!

Ber [7]

Answer: it’s the 3rd one

Step-by-step explanation:

7 0

4 years ago

Math help please and thank you

Doss [256]

for what? :) .

...................

7 0

3 years ago

Read 2 more answers