Application of first derivative: f(x) = sqrt[2x + 6]

[cmnalo]

I need to find the interval(s) where f(x) = square root (2x + 6)

f'(x) = 1/2(2x + 6)^-1/2

this where I get jammed up. In not sure my derivative and what step to take next.

The answer is:
Increasing: (-3,infinity)
Dreasing: never

 

[skeeter]

your derivative is incorrect ... remember the chain rule?

to finish the problem, look up the first derivative test in your text.

 

[stapel]

I need to find the interval(s) where f(x) = square root (2x + 6)

I'm sorry, but the above is not a complete sentence. You need to find the intervals where f(x) is or does what?

Thank you.

Eliz.

 

[cmnalo]

f(x) = square root (2x+6)
f(x) = (2x+6)^1/2
f"(x) = 1/2 (2x+6)^-1/2 (2)

I don't quiet see how to use the chain rule wouldn't I use the general power rule?

 

[tkhunny]

Where did that '2' come from on the end?

 

[cmnalo]

Isn't the general power rule?

h'(x)=d/dx[f(x)]^n = n[f(x)]^n-1 f(x)

 

[tkhunny]

You had better look that up again. It makes little sense but appears to be related to several things.

Power (n ≠ 0)

\(\displaystyle \L\,\frac{dx^{n}}{dx} = n*x^{n-1}\)

It is not about more complicated expressions, but it is part of other things.

Chain (n ≠ 0)

\(\displaystyle \L\,\frac{d[f(x)]^{n}}{dx} = n*[f(x)]^{n-1}*f'(x)\)