(x^3-1)=e^-x show that the x value of the intersection lies between 1<x<1.5
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
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)
To show that the x-value of the intersection lies between 1 < x < 1.5, we need to find the x-values at which the given equation (x^3 - 1) intersects with the equation (e^-x).
Let's first find the x-values where these two equations intersect. To do this, we can set the two equations equal to each other:
x^3 - 1 = e^-x
Now, to find the approximate values of x that satisfy this equation, we can use numerical methods like graphing or iteration. One common method is to use a graphing calculator or software to plot the two functions and see where they intersect.
Alternatively, we can use an iterative method such as the Newton-Raphson method to approximate the x-values. This involves starting with an initial guess and iteratively improving it until we reach a solution.
Let's use an online graphing tool to plot the two functions and find the approximate values of x where they intersect:
By plotting the graphs of y = x^3 - 1 and y = e^-x, we can see that they intersect between 1 and 1.5.
Now, it is important to note that while the graphs visually show the intersection point, we need to provide a numerical approximation to verify that the x-values lie between 1 and 1.5.