document.write( "Question 1173487: Consider the function give below:
\n" ); document.write( "f(x)=2x^4+ax^3-4x^2+bx-18
\n" ); document.write( "It is known that f(2)=4 and that f(1)=0
\n" ); document.write( "G. Determine the instantaneous rate of change when x =1
\n" ); document.write( "H. Estimate the turning points and then find the slope at these turning points. Are your estimation correct?
\n" ); document.write( "I. Which of these turning point you estimated are maximum and which are the minimum?
\n" ); document.write( "(please explain and show your working)
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Algebra.Com's Answer #798869 by Boreal(15235)\"\" \"About 
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f(x)=2x^4+ax^3-4x^2+bx-18
\n" ); document.write( "f(2)=4=32+8a-16+2b-18
\n" ); document.write( "so 6=8a+2b
\n" ); document.write( "f(1)=0=2+a-4+b-18
\n" ); document.write( "so 20=a+b
\n" ); document.write( "a=20-b
\n" ); document.write( "6=160-8b+2b
\n" ); document.write( "-154=-6b
\n" ); document.write( "b=25 2/3
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\n" ); document.write( "b=20-a
\n" ); document.write( "6=8a+40-2a
\n" ); document.write( "-34=6a
\n" ); document.write( "a=-5 2/3
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\n" ); document.write( "The function is f(x)=2x^4-(17/3)x^3-4x^2+(77/3)x-18
\n" ); document.write( "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.
\n" ); document.write( "f(1)=0 so equation of tangent line at (1, 0) is y=(26/3)x-26/3
\n" ); document.write( "set derivative equal to 0.
\n" ); document.write( "8x^3-17x^2-8x+(77/3)-18=0
\n" ); document.write( "try 0 and slope is 7 2/3 so no turning point
\n" ); document.write( "at x=-1 slope is still positive
\n" ); document.write( "but at x=-2 the slope is strongly negative
\n" ); document.write( "estimate the turning point (a local minimum) between x=-1 and x=-2 (it is at x=-1.19)
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