Respuesta :
Answer:
Step-by-step 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
[tex](\sin(x)) ' = \cos(x)[/tex]
[tex](\cos(x))' = -\sin(x)[/tex]
[tex](cf(x))' = cf'(x)[/tex] where c is a constant.
[tex](x^n)' = nx^{n-1}[/tex]
[tex](f+g)' = f'+g'[/tex] where f,g are differentiable.
[tex](c)' =0[/tex] where c is a constant.
[tex](f(g(x))' = f'(g(x)) \cdot g'(x)[/tex] (chain rule)
Case 1: f(x) = 2+3x+3.
Using the properties from above, we have
1. [tex] f'(x) = 0+3+0 = 3[/tex]
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: [tex]f(x) = 6x+\sin(3x) [/tex]
1. [tex]f'(x) = 6+3\cos(3x)[/tex]
2. We have the equation
[tex]6+3\cos(3x)=0[/tex]
which is equivalent to
[tex] \cos(3x) = -2[/tex]
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.
