To show that 'x^3 - 2x = 41' has a solution between 3 and 4 we need to substitute x = 3 and x = 4 into the equation.

When x = 3 we get: (3)^3 - 2*3 = 21
When x = 4 we get: (4)^3 - 2*4 = 56

Now notice when x = 3 we get an answer less than 41 and when x = 4 we get an answer greater than 41.

This means there must be a value of x between 3 and 4 which is equal to 41.

We now need to find this solution correct to 1 decimal place.

To do this we need to test different values of x. Since we know the answer is between 3 and 4 it makes sense to start with x = 3.5.

When x = 3.5 we get: (3.5)^3 - 2*3 = 35.875 too small

When x = 3.7 we get: (3.7)^3 - 2*3.7 = 43.253  too big

When x = 3.6 we get: (3.6)^3 - 2*3.6 = 39.456 too small

Now since 3.6 was too small and 3.7 was too big we know the solution is between these values.

To find our answer to 1 decimal place we have to try one more value in the middle of these, when x = (3.6+3.7)/2

that is when x = 3.65

When x = 3.65 we get: (3.65)^3 - 2*3.65 = 41.327125 too big

Finally since 3.65 is too big then we can say the answer is between 3.6 and 3.65. Hence x = 3.6 to 1 decimal place.