Question

find all local maximum and minimum points by the second derivative test when possible y=2+3×+3 (b) y=6×+sin3×​

Answer 1

Explanation:

Given a function f, whose derivatives are f' and f'', a value x is a critical point if f'(x) =0. A value x is a minimum of f if it is a critical point and f''(x) >0 and it is maximum if f''(x)<0. We will perfom the following steps:

1. Calculate the derivative f'.

2. Solve f'(x) =0.

3. Determine if the x value found in 2 is a minimum or a maximum using f''.

Recall the following properties of derivatives

`(\sin(x)) ' = \cos(x)`

`(\cos(x))' = -\sin(x)`

`(cf(x))' = cf'(x)` where c is a constant.

`(x^n)' = nx^(n-1)`

`(f+g)' = f'+g'` where f,g are differentiable.

`(c)' =0` where c is a constant.

`(f(g(x))' = f'(g(x)) \cdot g'(x)` (chain rule)

Case 1: f(x) = 2+3x+3.

Using the properties from above, we have

1.
`f'(x) = 0+3+0 = 3`

2. The equation f'(x)=0 where f'(x) = 3 has no solution.

3. Based on the previous result, f has no maximum nor minimum.

Case 2:
`f(x) = 6x+\sin(3x)`

1.
`f'(x) = 6+3\cos(3x)`

2. We have the equation

`6+3\cos(3x)=0`

which is equivalent to

`\cos(3x) = -2`

Recall that the cosine function only takes values in the set [-1,1]. So, this equation has no solution.

3. Based on the previous result, f has no maximum nor minimum.