Are you talking about the intersection of
y = x^3 - 1 and y = e^-x ?
if so, the graph shows that they intersect here:
https://www.wolframalpha.com/input/?i=plot+%28x%5E3-1%29%3De%5E-x
here is the solution to your equation.
https://www.wolframalpha.com/input/?i=%28x%5E3-1%29%3De%5E-x
which is between 1 and 1.5
(x^3-1)=e^-x show that the x value of the intersection lies between 1<x<1.5
2 answers
If (x^3-1)=e^-x
then f(x) = (x^3-1) - e^-x = 0
f(1) = 0 - 1 = -1 < 0
f(1.5) = 1.25 - 0.37 = 0.88 > 0
Since f(x) is continuous, it must be zero in the interval (1,1.5)
then f(x) = (x^3-1) - e^-x = 0
f(1) = 0 - 1 = -1 < 0
f(1.5) = 1.25 - 0.37 = 0.88 > 0
Since f(x) is continuous, it must be zero in the interval (1,1.5)
1 answer