SOLUTION: Consider the function give below:
f(x)=2x^4+ax^3-4x^2+bx-18
It is known that f(2)=4 and that f(1)=0
G. Determine the instantaneous rate of change when x =1
H. Estimate the turn
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Question 1173487: Consider the function give below:
f(x)=2x^4+ax^3-4x^2+bx-18
It is known that f(2)=4 and that f(1)=0
G. Determine the instantaneous rate of change when x =1
H. Estimate the turning points and then find the slope at these turning points. Are your estimation correct?
I. Which of these turning point you estimated are maximum and which are the minimum?
(please explain and show your working)
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
f(x)=2x^4+ax^3-4x^2+bx-18
f(2)=4=32+8a-16+2b-18
so 6=8a+2b
f(1)=0=2+a-4+b-18
so 20=a+b
a=20-b
6=160-8b+2b
-154=-6b
b=25 2/3
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b=20-a
6=8a+40-2a
-34=6a
a=-5 2/3
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The function is f(x)=2x^4-(17/3)x^3-4x^2+(77/3)x-18
When x=1 f'(x)=8x^3-17x^2-8x+(77/3) and f(1)=8-17-8+77/3=8 2/3 units, instantaneous rate of change.
f(1)=0 so equation of tangent line at (1, 0) is y=(26/3)x-26/3
set derivative equal to 0.
8x^3-17x^2-8x+(77/3)-18=0
try 0 and slope is 7 2/3 so no turning point
at x=-1 slope is still positive
but at x=-2 the slope is strongly negative
estimate the turning point (a local minimum) between x=-1 and x=-2 (it is at x=-1.19)
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