Instantaneous Rate of Change Problems!

Question

Let g(x)=2x2+4x+1.
Find a value of c between 1 and 5 such that the average rate of change of g(x) from x=1 to x=5 is equal to the instantaneous rate of g(x) at x=c.

 

Suppose h(t)=t2+14t+7. Find the instantaneous rate of change of h(t) with respect to t at t=2.
The instantaneous rate of change is

 

Suppose G(x)=3x2+x+4. Find a number b such that G′(b)=5.

 

Let F(s)=5s2+3s+4.
Find a value of d greater than 0 such that the average rate of change of F(s) from 0 to d equals the instantaneous rate of change of F(s) at s=4.

 

Let f(t)=6t2+3t+1.
Find the value of t for which the tangent line to the graph of f(t) has slope 1.

 

Let f(x)=x2+5x+13.
What is the value of x for which the tangent line to the graph of y=f(x) is parallel to the x-axis?

 

Let f(x)=2x2−4x+10.
What is the value of x for which the tangent line to the graph of y=f(x) is parallel to the x-axis?

 

A particle is traveling along a straight line. Its position at time t is given by s(t)=5t2+7. Find the velocity at time t=3.
The velocity at t=3 is

 

If h(t) represents the height of an object above ground level at time t and h(t) is given by
h(t)=−16t2+11t+13
find the speed at time t=0.
The speed at time t=0 equals

 

If h(t) represents the height of an object in feet above ground level at time t and h(t) is given by
h(t)=−16t2+13t+3
find the height of the object at the time when the speed is zero.
The height of the object is h(t)=

Answer 1

16 = average rate of change between x =1 and x=5

= (g(5)-g(1))/(5-1) = (50+20+1 -2-4-1)/4 = 64/4 = 16

g(x) = 2x^2 +4x +1 has derivative = g'(x) = 4x +4 = 16, x+1 =4, x = 3 = c

f(c) = f(3) = 16 = g'(3) = g"(x) = g'(c)

let x = s in next problem

F(x) = 5X^2 + 3x + 4 find d such that average rate of change from x= 0 to d is same as the instantaneous rate of change at x =4

F'(4 = 10(4) +3(4) = 52 = (F(d) - F(0))/(d -0) = 5d^2 +3d)/d = 5d +3 =52, d = 49/5

s(t) = 5t^2 + 1

v(t) = s'(t) = 10t

v(3) = 10(3) = 30 = velocity at time t=3

f(x) = x^2 +5x +13 has tangent line parallel to the x axis when

f'(x) = 0 = 2x +5, x =-5/2

the point (-5/2, f(-5/2) = (-5/2, 77/4)

f(x) = 2x^2 -4x + 10

f'(x) = 0 = 4x-4, x =1 where tangent line is parallel to the x axis

h(t) =-16t^2 +11t +13

has velocity = h'(t) = -32t +11

h'(0) = 11 feet/second = velocity at time t=0

velocity = 0 when -32t+11 = 0, at time t = 11/32 = about 1/3 of a second

h(11/32) = -16(11/32)^2 +11(11/32) +13 = -121/64 + 242/64 +13 = 13+121/64

= 13(64)/64 + 121/64 = (833+121)/64 = 954/64 = 477/32 = 14.90625 feet high = the maximum height