My calculator gave me a negative value for f(1.6)
and a positive value for f(1.8)
I wonder how the graph got from below the x-axis to above the x-axis in that little interval.
Use the intermediate theorem to show that the polynomial function value has a zero in the given interval
f(x)=x^5-x^4+8x^3-7x^2-17x+7; [1.6,1.8]
Find the value of f(1.6)
Find the value of f(1.8)
3 answers
What does that mean?
mmmhh?
I plugged in x = 1.6 and got some negative value
I plugged in x = 1.8 and got some positive value
so we have two points of your graph
one is (1.6, -?) , the other is (1.8, + ?)
so one is below the x-axis , the other is above the x-axis
make a rough sketch of this situation, (it doesn't make any difference how far down or how far up you place points)
join your two points.
I see them crossing or intersecting the x-axis.
Don't we call the place where a graph cuts the x-axis, "the zero of the function" ??
I see such a point between x = 1.6 and 1.8
I plugged in x = 1.6 and got some negative value
I plugged in x = 1.8 and got some positive value
so we have two points of your graph
one is (1.6, -?) , the other is (1.8, + ?)
so one is below the x-axis , the other is above the x-axis
make a rough sketch of this situation, (it doesn't make any difference how far down or how far up you place points)
join your two points.
I see them crossing or intersecting the x-axis.
Don't we call the place where a graph cuts the x-axis, "the zero of the function" ??
I see such a point between x = 1.6 and 1.8
1 answer