document.write( "Question 1173487: Consider the function give below:
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document.write( "f(x)=2x^4+ax^3-4x^2+bx-18
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document.write( "It is known that f(2)=4 and that f(1)=0
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document.write( "G. Determine the instantaneous rate of change when x =1
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document.write( "H. Estimate the turning points and then find the slope at these turning points. Are your estimation correct?
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document.write( "I. Which of these turning point you estimated are maximum and which are the minimum?
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document.write( "(please explain and show your working) \n" );
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Algebra.Com's Answer #798869 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! 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 \n" ); document.write( "- \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 \n" ); document.write( "- \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) \n" ); document.write( "- \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |