# Is #f(x)=sqrt(x+2) # increasing or decreasing at #x=2 #?

By signing up, you agree to our Terms of Service and Privacy Policy

To determine if ( f(x) = \sqrt{x + 2} ) is increasing or decreasing at ( x = 2 ), we evaluate the derivative of ( f(x) ) at ( x = 2 ).

The derivative of ( f(x) ) with respect to ( x ) is: [ f'(x) = \frac{1}{2\sqrt{x + 2}} ]

Evaluating this derivative at ( x = 2 ): [ f'(2) = \frac{1}{2\sqrt{2 + 2}} = \frac{1}{2\sqrt{4}} = \frac{1}{4} ]

Since ( f'(2) > 0 ), the function is increasing at ( x = 2 ).

By signing up, you agree to our Terms of Service and Privacy Policy

- What are the critical values, if any, of #f(x)=x^(4/5) (x − 3)^2#?
- How do you find the critical points and the open intervals where the function is increasing and decreasing for #y = xe^(x(2 - 3x))#?
- What are the global and local extrema of #f(x)=x^3-x^2-x+1# ?
- How do use the first derivative test to determine the local extrema #f(x)=x-2tan(x)#?
- How do you determine all values of c that satisfy the mean value theorem on the interval [2,5] for #f(x) = 1 / (x-1)#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7