Question

Show that f'(x)=0 at some point between the two x-intercepts of the function

f(x) = -6x√x+1

x intercepts (-1,0) and (0,0)

Find a value of x such that f'(x) = 0

x=

Answer 1

This is calc 1 right?

Answer 2

yes

Answer 3

Not really sure about the first part but the way you find value for x such that f'(x)=0 by taking the derivative of the given function and then setting it to zero and solving for x

Answer 4

I'm thinking for the first part you can plug in the intercepts and take the derivative of them to show that f;(x) is in between

Answer 5

\[-6x \sqrt{x+1} = -6x[\frac{ 1 }{ 2 }(x+1)]^{-1/2}\]

Answer 6

Actually there is only one part, finding the value of x

Answer 7

Well it is asking you to show something, and then it asks you to find something.

Answer 8

I think they mean the same thing

Answer 9

Is this a problem in a book?

Answer 10

From an online course. The original question was:
Find the two x-intercepts of the function f and show that
f '(x) = 0 at some point between the two x-intercepts.

The second part of the question was:
Find a value of x such that
f '(x) = 0.

Answer 11

Okay, for the second part I believe you take the derivative of f(x) and then set f'(x)=0. Then solve for x.

Answer 12

\[x*\sqrt{x}=x^{3/2}\]

Answer 13

\[f'(x)=-6*\frac{ 3 }{ 2 }*x^{1/2}+0\]

Answer 14

f'(x)=0=-9sqrtx=0

Answer 15

x=0

Answer 16

I got x=-2/3

Answer 17

\[f'(x)=-6\sqrt{x+1}-\frac{ 3x }{ \sqrt{x+1} }\]

Answer 18

by the product rule

Answer 19

I'm confused at where that 3 comes from

Answer 20

Has your teacher covered the product rule yet?

Answer 21

yes. we've been through that

Answer 22

Derivative of the first times the second plus the first times the derivative of the second

Answer 23

actually it's an all online course so I have to learn it myself

Answer 24

Essentially this: \[-6(x+1)^\frac{ 1 }{ 2 } - (6x)(\frac{ 1 }{ 2 })(x+1)^\frac{- 1 }{ 2 }\]

Answer 25

I took the derivative of the first term multiplied it by the second term

Answer 26

vice versa for the other half

Answer 27

\[-6x(x+1)^{-1/2}(\frac{ 3 }{ 2 }x+1)=0\]

Answer 28

http://www.derivative-calculator.net/#expr=-6x%28x%2B1%29%5E%281%2F2%29&showsteps=1

Answer 29

WolframAlpha is also good

Answer 30

Yeah it is.

Answer 31

\[-\frac{ 3 (3x+2) }{ \sqrt{x+1} }\]

Answer 32

Is that the simplified form?

Answer 33

yes.

Answer 34

it comes from \[-6(\sqrt{1+x}+1\frac{ x }{ 2\sqrt{1+x}})\]

Answer 35

I'm not sure how -2/3 comes out of that

Answer 36

Let me solve real quick

Answer 37

\[f'(x)=-6\sqrt{x+1}-\frac{ 3x }{ \sqrt{x+1} }\]
\[0=-6\sqrt{x+1}-\frac{ 3x }{ \sqrt{x+1} }\]
\[0=-6(x+1)-3x\]
\[0=-6x-6-3x\]
\[9x=-6\]

Answer 38

\[x=-\frac{ 6 }{ 9 }=-\frac{ 2 }{ 3 }\]

Answer 39

ok. think I've got it now

Answer 40

Okay great. Good luck!