FINDING INSTANTANEOUS RATE OF CHANGE AT A POINT

Instantaneous rate of change :

The instantaneous rate of change of a function f(x) at x = a is given by 

Instantaneous rate of change = lim h → 0 f(x + h) - f(x)h

For each problem, find the instantaneous rate of change of the function at the given value.

Problem 1 :

y = x2 – 2x + 1; 2

A)  0

C)  4

B)  -1/2

D)  2

Solution :

 y = f(x)

Given, f(x) =  x2 – 2x + 1

Instantaneous rate of change = lim h → 0 f(x + h) - f(x)hx = 2= lim h → 0 f(2 + h) - f(2)hf(2) = (2)2 - 2(2) + 1= 4 - 4 + 1f(2) = 1f(2 + h) = (2 + h)2 - 2(2 + h) + 1= 22 + h2 + 2(2)(h) - 4 - 2h + 1= 4 + h2 + 4h - 4 - 2h + 1f(2 + h)= h2 + 2h + 1= lim h → 0 h2 + 2h + 1 - 1h= lim h → 0 h2 + 2h h= lim h → 0 h(h + 2) h = lim h → 0 h + 2= 2

So, the instantaneous rate of change of the function is 2.

Hence, option D) is correct.

Problem 2 :

f(x) = x2 + 2x + 2; -2

A)  -1/2

C)  -2

B)  1

D)  8

Solution :

Given, f(x) =  x2 + 2x + 2

Instantaneous rate of change = lim h → 0 f(x + h) - f(x)hx = -2= lim h → 0 f(-2 + h) - f(-2)hf(-2) = (-2)2 + 2(-2) + 2= 4 - 4 + 2f(-2) = 2f(-2 + h) = (-2 + h)2 + 2(-2 + h) + 2= -22 + h2 + 2(-2)(h) - 4 + 2h + 2= 4 + h2 - 4h - 4 + 2h + 2f(-2 + h)= h2 - 2h - 2= lim h → 0 h2 - 2h - 2 + 2h= lim h → 0 h2 - 2h h= lim h → 0 h(h - 2) h = lim h → 0 h - 2= -2

So, the instantaneous rate of change of the function is -2.

Hence, option C) is correct.

Problem 3 :

y = -2x2 + x + 1; 0

A)  -2

C)  1/4

B)  1

D)  -1/2

Solution :

y = f(x)

Given, f(x) =  -2x2 + x + 1

Instantaneous rate of change = lim h → 0 f(x + h) - f(x)hx = 0= lim h → 0 f(0 + h) - f(0)hf(0) = -2(0)2 + 0 + 1f(0) = 1f(0 + h) = -2(0 + h)2 + 0 + h + 1= -202 + h2 + 2(0)(h) + h + 1f(0 + h)= -2h2 + h + 1= lim h → 0 -2h2 + h + 1 - 1h= lim h → 0 -2h2 + h h= lim h → 0 h(-2h + 1)h = lim h → 0 (-2h + 1)= 1

So, the instantaneous rate of change of the function is 1.

Hence, option B) is correct.

Problem 4 :

y = 2x2 + 1; 0

A)  1/3

C)  -3

B)  4

D)  0

Solution :

y = f(x)

Given, f(x) =  2x2 + 1

Instantaneous rate of change = lim h → 0 f(x + h) - f(x)hx = 0= lim h → 0 f(0 + h) - f(0)hf(0) = 2(0)2 + 1f(0) = 1f(0 + h) = 2(0 + h)2 + 1= 202 + h2 + 2(0)(h) + 1f(0 + h)= 2h2 + 1= lim h → 0 2h2 + 1 - 1h= lim h → 0 2h2 h= lim h → 0 2h = lim h → 0 (2(0)) = 0

So, the instantaneous rate of change of the function is 0.

Hence, option D) is correct.

Problem 5 :

y =-1x - 1 ; -2
A) 13
C) 49
B) -13
D) 19

Solution :

Given, y =-1x - 1 ; -2Instantaneous rate of change = lim h → 0 f(x + h) - f(x)hx = 0= lim h → 0 f(-2 + h) - f(-2)hf(-2) = -1-2 - 1 f(-2) = -1-3 f(-2) = 13 f(-2 + h) = -1-2 + h - 1 =-1-3 + h = lim h → 0 -1-3 + h - 13h= lim h → 0 13 - h - 13h= lim h → 0 3 - (3 - h)3(3 - h) h= lim h → 0 3 - 3 + h9 - h h= lim h → 0 h9 - h × 1h = lim h → 0 19 - h= 19

So, the instantaneous rate of change of the function is 1/9.

Hence, option D) is correct.

Problem 6 :

y =1x + 2 ; -1

      A)  -1

C) 14
B) 13
D) 12

Solution :

Given, y =1x + 2 ; -1Instantaneous rate of change = lim h → 0 f(x + h) - f(x)hx = -1= lim h → 0 f(-1 + h) - f(-1)hf(-1) = 1-1 + 2 f(-1) = 11 f(-1) = 1f(-1 + h) = 1-1 + h + 2 =-11 + h = lim h → 0 -11 + h - 1h= lim h → 0 1 - (1 + h)1 + hh= lim h → 0 1 - 1 - h1 + h h= lim h → 0 -h 1 + h × 1h = lim h → 0 -11 + h= -1

So, the instantaneous rate of change of the function is -1.

Hence, option A) is correct.

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