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Llana [10]
3 years ago
9

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].
A. 0
B. 2.5
C. 4.5
D. 11.5

Mathematics
1 answer:
ivolga24 [154]3 years ago
4 0

Answer:

B

Step-by-step explanation:

f(0) - f(-3) = area under f'(x) from x=0 to x=3.

We can find the area under f'(x) in this interval by finding the area of the triangle formed by the line.

A = 1/2 b h

A = 1/2 (3) (3)

A = 4.5

f(0) - f(-3) = 4.5

Since f(0) = 7:

7 - f(-3) = 4.5

f(-3) = 7 - 4.5

f(-3) = 2.5

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Determine the value of c so that f(x) is continuous on the entire real line when
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Answer:

A. -4

Step-by-step explanation:

Given the function f(x) = x + 3 for x ≤ -1 and 2x - c for x > -1, for the function to be continuous, the right hand limit of the function must be equal to its left hand limit.

For the left hand limit;

The function at the left hand occurs at x<-1

f-(x) = x+3

f-(-1) = -1+3

f-(-1) = 2

For the right hand limit, the function occurs at x>-1

f+(x) = 2x-c

f+(-1) = 2(-1)-c

f+(-1) = -2-c

For the function f(x) to be continuous on the entire real line at x = -1, then

f-(-1) = f+(-1)

On equating both sides:

2 = -2-c

Add 2 to both sides

2+2 = -2-c+2

4 =-c

Multiply both sides by minus.

-(-c) = -4

c = -4

Hence the value of c so that f(x) is continuous on the entire real line is -4

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Answer:

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