# Please help: find the critical values and relative extrema for the function on the attached image.

*print*Print*list*Cite

### 1 Answer

`f(x) = -4x^2 + 24x+ 2`

a) Critical values are the values where the derivative is zero and does not exist. The first derivative of the function is

f'(x) = -8x + 24

This exists everywhere and is 0 when -8x + 24 = 0.

-8x = -24

x = 3

**The critical value is 3.**

b) The function is increasing when the first derivative is positive. This happens when -8x + 24 >0

From here -8x > -24. Divide by -8, and change the sign of inequality:

x < 3.

In interval notation, it is written as `(-oo, 3)` .

The interval when function is decreasing can be found in a similar way. The function is decreasing when the fist derivative is negative:

-8x + 24 <0

-8x < -24

x > 3

The function is decreasing on the interval

`(3, oo)` .

c) According to the First Derivative Test, if the function's derivative changes sign at a critical value, there is a local (relative) extremum at this value. The given function has one critical value x = 3, where f'(x) changes sign from positive to negative. This means f(x) has a relative maximum at x = 3.

At x = 3, `y = f(3) = -4*3^2 + 24*3 + 2 = 38` .

**The relative maximum is (3, 38).**

**The relative minimum does not exist.**