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Suppose f is a function whose derivative f' is given by the graph shown next. In Exercises 9–12, use the given value of f , an area approximation, and the Fundamental Theorem to approximate the requested value

Given that f(-1)=2, approximate f(1).

Short Answer

Expert verified

The required approximationf(1)=5.17

Step by step solution

01

Step 1. Given information

Given that f(-1)=2and graph of f'

02

Step 2. Finding the value of f(1)

Use the method. If fis derivative of the function f'then they by fundamental theorem f(x)=∫f'(x)dx+C

The graph (parabola) of the function shows that f'is quadratic function whose roots are 2.5 and -0.5

Therefore,f'(x)=(x-2.5)(x-0.5)=x2-3x+1.25Now,f(x)=f'(x)dx+C,whereCisaconstant

⇒f(x)=∫x2-3x+1.25dx+C⇒f(x)=x33-3x22+1.25x+C............(1)

we have f(-1)=2...........(2)

From results (1) and (2)

f(-1)=-133-3(-1)22+1.25(-1)+C⇒f(-1)=-13-32-54+C⇒2=-13-32-54+CC=2+13+32+54=24+4+18+1512=6112

Substitute value of C in result (1)

f(x)=x33-3x22+54x+6112.......................(3)

f(1)=(1)33-3122+54(1)+6112=13-32+54+6112⇒f(1)=4-18+15+6112=6212=5.17

Thus,f(1)=5.17

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