Question

The graph of f ′ (x), the derivative of f of x, is continuous for all x and consists of five line segments as shown below. Given f (0) = 7, find the absolute minimum value of f (x) over the interval [–3, 0].

Answer 1

Since x=0 is the value of x for which f'(x)=0, f(0) gives the value of f(x) at which it is either maximum or minimum thus the value 7 is either it's maximum or minimum. If f'(x) changes from negative to positive ie, f'(x)>0 for some x<c and f'(x)<0 for some x>c, then it's a point of local maxima here the c is 0, it's the value of x at which f'(x) is 0 so take any value less than 0 and more than 0 and look at f'(x) and if it's f'(x)<0 and f'(x)>0 for x<c and x>c respectively then it's minima

Answer 2

Answer:

Absolute minimum = 7

Step-by-step explanation:

We are given graph of f'(x). f(0)=7

We need to find the absolute minimum value of f(x) over interval [-3,0]

First we will see the graph of f'(x) over interval [-3.0]

f'(-3)=3

f'(0)=0

Thus, f'(x) is decreasing

x=0 is critical point of the function f(x) because f'(0)=0

We will get absolute maximum/minimum at x=0. f(x) >0

Hence, f(0) is absolute minimum at x=0 , Absolute minimum = 7