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

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

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

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

$(cf(x))' = cf'(x)$ 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.