Help me find the rate of change

F(x)=(x+1)^(5)-5x-2 <br> Find F'(x) and zeros of F'(x)

KiRa [710]

Assuming you mean
$f(x)=(x+1)^5-5x-2$

first use chain rule on first part
derivitive of (x+1)⁵=5(x+1)⁴ times the derivitive of (x+1) which is 1
the first part is 5(x+1)⁴

derivitive of -5x is -5
derivitive of -2 is 0

we gots
f'(x)=5(x+1)⁴-5
find zeroes
0=5(x+1)⁴-5
5=5(x+1)⁴
1=(x+1)⁴
+/-1=x+1
-1+/-1=x
-2=x or 0=x

f'(x)=5(x+1)⁴-5
zeroes at x=-2 and x=0

3 0

3 years ago

Carmen paid $10.50 for a T-shirt at the mall. It was on sale for 30% off. What was the original price before the discount? *

GREYUIT [131]

Answer:

15

Step-by-step explanation:

70%-> 10.50

100%-> 15

3 0

3 years ago

Please put down the steps?

e-lub [12.9K]

The answer is: " x = 0, 1 " .
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Explanation:
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Given:
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" √(x + 1) <span>− 1 = x " ; Solve for "x" ;

First, let us assume that "x </span>≥ -1 "
<span>
Add "1" to EACH SIDE of the equation:

</span>→ √(x + 1) − 1 + 1 = x + 1 ;

to get:

→ √(x + 1) = x + 1 .

Now, "square" EACH side of the equation:

→ [√(x + 1) ]² = (x + 1 )² ;

to get:

x + 1 = (x + 1)²

↔ (x + 1)² = (x + 1) .

Expand the "left-hand side" of the equation:

→ (x + 1)² = (x + 1)(x +1) ;

Note: (a+b)(c+d) = ac +ad + bc + bd ;

As such: (x + 1)(x + 1) = (x*x) + (x*1) +(1(x) + (1*1) ;
= x² + 1x + 1x + 1 ;
= x² + 2x + 1 ;

Now, substitute this "expanded" value, and bring down the "right-hand side" of the equation; and rewrite the equation:
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" (x + 1)² = (x + 1)(x +1) " ;

→ Rewrite as: " x² + 2x + 1 = x + 1 " ;

Subtract "x" ; and subtract "1" ; from EACH SIDE of the equation:

→ x² + 2x + 1 - x - 1 = x + 1 - x - 1 ;

to get: → x² <span>− x = 0

Factor out an "x" on the "left-hand side" of the equation:
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</span>x² − x = x(x − 1) ;

→ x (x − 1) = 0 ;

We have: "x" and "(x − 1)" ; when either of these two multiplicands are equal to zero, then the "right-hand side of the equation equals "zero" .

So, one value of "x" is "0" .

The other value for "x" ;

→ x − 1 = 0 ;

Add "1" to each side of the equation:

→ x − 1 + 1 = 0 + 1 ;

→ x = 1 ;
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So, the answers:

" x = 0, 1 " .
_________________________________

4 0

4 years ago

Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one sam

Makovka662 [10]

<h2>Steps</h2>

So for this, we will create a system of equations. With our information given, these are our two equations (let x = advance and y = same-day):

$x+y=65\\20x+30y=1700$