EDU>> lesson1
Solution to exercise problem
Demonstration of tangents at a point
for the equation y=f(x)=x^2
where slope m (at point a) = ( f(a+h) - f(a) ) / h

At what point a is tangent desired (enter value from 0 to 12)?10
for h=1.000000 : then ( f(10.000000+1.000000)-f(10.000000) )/ 1.000000) = ( 121.000000-100.000000)/ 1.000000) = 21.000000
for h=0.500000 : then ( f(10.000000+0.500000)-f(10.000000) )/ 0.500000) = ( 110.250000-100.000000)/ 0.500000) = 20.500000
for h=0.250000 : then ( f(10.000000+0.250000)-f(10.000000) )/ 0.250000) = ( 105.062500-100.000000)/ 0.250000) = 20.250000
for h=0.125000 : then ( f(10.000000+0.125000)-f(10.000000) )/ 0.125000) = ( 102.515625-100.000000)/ 0.125000) = 20.125000
for h=0.062500 : then ( f(10.000000+0.062500)-f(10.000000) )/ 0.062500) = ( 101.253906-100.000000)/ 0.062500) = 20.062500
for h=0.031250 : then ( f(10.000000+0.031250)-f(10.000000) )/ 0.031250) = ( 100.625977-100.000000)/ 0.031250) = 20.031250
for h=0.015625 : then ( f(10.000000+0.015625)-f(10.000000) )/ 0.015625) = ( 100.312744-100.000000)/ 0.015625) = 20.015625
for h=0.007813 : then ( f(10.000000+0.007813)-f(10.000000) )/ 0.007813) = ( 100.156311-100.000000)/ 0.007813) = 20.007813
for h=0.003906 : then ( f(10.000000+0.003906)-f(10.000000) )/ 0.003906) = ( 100.078140-100.000000)/ 0.003906) = 20.003906
for h=0.001953 : then ( f(10.000000+0.001953)-f(10.000000) )/ 0.001953) = ( 100.039066-100.000000)/ 0.001953) = 20.001953
for h=0.000977 : then ( f(10.000000+0.000977)-f(10.000000) )/ 0.000977) = ( 100.019532-100.000000)/ 0.000977) = 20.000977
EDU>> 
EDU>> lesson1
Solution to exercise problem
Demonstration of tangents at a point
for the equation y=f(x)=10x
where slope m (at point a) = ( f(a+h) - f(a) ) / h

At what point a is tangent desired (enter value from 0 to 12)?2
for h=1.000000 : then ( f(2.000000+1.000000)-f(2.000000) )/ 1.000000) = ( 30.000000-20.000000)/ 1.000000) = 10.000000
for h=0.500000 : then ( f(2.000000+0.500000)-f(2.000000) )/ 0.500000) = ( 25.000000-20.000000)/ 0.500000) = 10.000000
for h=0.250000 : then ( f(2.000000+0.250000)-f(2.000000) )/ 0.250000) = ( 22.500000-20.000000)/ 0.250000) = 10.000000
for h=0.125000 : then ( f(2.000000+0.125000)-f(2.000000) )/ 0.125000) = ( 21.250000-20.000000)/ 0.125000) = 10.000000
for h=0.062500 : then ( f(2.000000+0.062500)-f(2.000000) )/ 0.062500) = ( 20.625000-20.000000)/ 0.062500) = 10.000000
for h=0.031250 : then ( f(2.000000+0.031250)-f(2.000000) )/ 0.031250) = ( 20.312500-20.000000)/ 0.031250) = 10.000000
for h=0.015625 : then ( f(2.000000+0.015625)-f(2.000000) )/ 0.015625) = ( 20.156250-20.000000)/ 0.015625) = 10.000000
for h=0.007813 : then ( f(2.000000+0.007813)-f(2.000000) )/ 0.007813) = ( 20.078125-20.000000)/ 0.007813) = 10.000000
for h=0.003906 : then ( f(2.000000+0.003906)-f(2.000000) )/ 0.003906) = ( 20.039063-20.000000)/ 0.003906) = 10.000000
for h=0.001953 : then ( f(2.000000+0.001953)-f(2.000000) )/ 0.001953) = ( 20.019531-20.000000)/ 0.001953) = 10.000000
for h=0.000977 : then ( f(2.000000+0.000977)-f(2.000000) )/ 0.000977) = ( 20.009766-20.000000)/ 0.000977) = 10.000000
EDU>> 
EDU>> lesson1
Solution to exercise problem
Demonstration of tangents at a point
for the equation y=f(x)=1
where slope m (at point a) = ( f(a+h) - f(a) ) / h

At what point a is tangent desired (enter value from 0 to 12)?6
for h=1.000000 : then ( f(6.000000+1.000000)-f(6.000000) )/ 1.000000) = ( 1.000000-1.000000)/ 1.000000) = 0.000000
for h=0.500000 : then ( f(6.000000+0.500000)-f(6.000000) )/ 0.500000) = ( 1.000000-1.000000)/ 0.500000) = 0.000000
for h=0.250000 : then ( f(6.000000+0.250000)-f(6.000000) )/ 0.250000) = ( 1.000000-1.000000)/ 0.250000) = 0.000000
for h=0.125000 : then ( f(6.000000+0.125000)-f(6.000000) )/ 0.125000) = ( 1.000000-1.000000)/ 0.125000) = 0.000000
for h=0.062500 : then ( f(6.000000+0.062500)-f(6.000000) )/ 0.062500) = ( 1.000000-1.000000)/ 0.062500) = 0.000000
for h=0.031250 : then ( f(6.000000+0.031250)-f(6.000000) )/ 0.031250) = ( 1.000000-1.000000)/ 0.031250) = 0.000000
for h=0.015625 : then ( f(6.000000+0.015625)-f(6.000000) )/ 0.015625) = ( 1.000000-1.000000)/ 0.015625) = 0.000000
for h=0.007813 : then ( f(6.000000+0.007813)-f(6.000000) )/ 0.007813) = ( 1.000000-1.000000)/ 0.007813) = 0.000000
for h=0.003906 : then ( f(6.000000+0.003906)-f(6.000000) )/ 0.003906) = ( 1.000000-1.000000)/ 0.003906) = 0.000000
for h=0.001953 : then ( f(6.000000+0.001953)-f(6.000000) )/ 0.001953) = ( 1.000000-1.000000)/ 0.001953) = 0.000000
for h=0.000977 : then ( f(6.000000+0.000977)-f(6.000000) )/ 0.000977) = ( 1.000000-1.000000)/ 0.000977) = 0.000000
EDU>> 
EDU>> lesson1
Solution to exercise problem
Demonstration of tangents at a point
for the equation y=f(x)=x^2+x+1
where slope m (at point a) = ( f(a+h) - f(a) ) / h

At what point a is tangent desired (enter value from 0 to 12)?2
for h=1.000000 : then ( f(2.000000+1.000000)-f(2.000000) )/ 1.000000) = ( 13.000000-7.000000)/ 1.000000) = 6.000000
for h=0.500000 : then ( f(2.000000+0.500000)-f(2.000000) )/ 0.500000) = ( 9.750000-7.000000)/ 0.500000) = 5.500000
for h=0.250000 : then ( f(2.000000+0.250000)-f(2.000000) )/ 0.250000) = ( 8.312500-7.000000)/ 0.250000) = 5.250000
for h=0.125000 : then ( f(2.000000+0.125000)-f(2.000000) )/ 0.125000) = ( 7.640625-7.000000)/ 0.125000) = 5.125000
for h=0.062500 : then ( f(2.000000+0.062500)-f(2.000000) )/ 0.062500) = ( 7.316406-7.000000)/ 0.062500) = 5.062500
for h=0.031250 : then ( f(2.000000+0.031250)-f(2.000000) )/ 0.031250) = ( 7.157227-7.000000)/ 0.031250) = 5.031250
for h=0.015625 : then ( f(2.000000+0.015625)-f(2.000000) )/ 0.015625) = ( 7.078369-7.000000)/ 0.015625) = 5.015625
for h=0.007813 : then ( f(2.000000+0.007813)-f(2.000000) )/ 0.007813) = ( 7.039124-7.000000)/ 0.007813) = 5.007813
for h=0.003906 : then ( f(2.000000+0.003906)-f(2.000000) )/ 0.003906) = ( 7.019547-7.000000)/ 0.003906) = 5.003906
for h=0.001953 : then ( f(2.000000+0.001953)-f(2.000000) )/ 0.001953) = ( 7.009769-7.000000)/ 0.001953) = 5.001953
for h=0.000977 : then ( f(2.000000+0.000977)-f(2.000000) )/ 0.000977) = ( 7.004884-7.000000)/ 0.000977) = 5.000977
EDU>> 
EDU>> lesson1
Solution to exercise problem
Demonstration of tangents at a point
for the equation y=f(x)=x^3-5x^2+6x+3
where slope m (at point a) = ( f(a+h) - f(a) ) / h

At what point a is tangent desired (enter value from 0 to 12)?1
for h=1.000000 : then ( f(1.000000+1.000000)-f(1.000000) )/ 1.000000) = ( 3.000000-5.000000)/ 1.000000) = -2.000000
for h=0.500000 : then ( f(1.000000+0.500000)-f(1.000000) )/ 0.500000) = ( 4.125000-5.000000)/ 0.500000) = -1.750000
for h=0.250000 : then ( f(1.000000+0.250000)-f(1.000000) )/ 0.250000) = ( 4.640625-5.000000)/ 0.250000) = -1.437500
for h=0.125000 : then ( f(1.000000+0.125000)-f(1.000000) )/ 0.125000) = ( 4.845703-5.000000)/ 0.125000) = -1.234375
for h=0.062500 : then ( f(1.000000+0.062500)-f(1.000000) )/ 0.062500) = ( 4.929932-5.000000)/ 0.062500) = -1.121094
for h=0.031250 : then ( f(1.000000+0.031250)-f(1.000000) )/ 0.031250) = ( 4.966827-5.000000)/ 0.031250) = -1.061523
for h=0.015625 : then ( f(1.000000+0.015625)-f(1.000000) )/ 0.015625) = ( 4.983891-5.000000)/ 0.015625) = -1.031006
for h=0.007813 : then ( f(1.000000+0.007813)-f(1.000000) )/ 0.007813) = ( 4.992066-5.000000)/ 0.007813) = -1.015564
for h=0.003906 : then ( f(1.000000+0.003906)-f(1.000000) )/ 0.003906) = ( 4.996063-5.000000)/ 0.003906) = -1.007797
for h=0.001953 : then ( f(1.000000+0.001953)-f(1.000000) )/ 0.001953) = ( 4.998039-5.000000)/ 0.001953) = -1.003902
for h=0.000977 : then ( f(1.000000+0.000977)-f(1.000000) )/ 0.000977) = ( 4.999022-5.000000)/ 0.000977) = -1.001952
EDU>>