Work out the x-coordinates on the curve y = x^5 - x where the gradient is 79?

If you have a curve y = f(x), the function for the gradient of the curve will be f'(x) (differential of the function)

we have y = x^5 - x
=> gradient of the curve = (x^5 - x) dx
=> gradient of the curve = 5x^4 - 1

We have to find the x coordinate at which gradient is 79

=> 79 = 5x^4 - 1
=> 80 = 5x^4
=> 16 = x^4
=> x^4 - 16 = 0
=> (x^2 + 4)(x + 2)(x - 2) = 0

x^2 + 4 is always greater than 0 (in other words, its always positive), so we can divide both sides of the equation by (x^2 + 4)
 
=> (x + 2)(x - 2) = 0
=> x + 2 = 0; x - 2 = 0
=> x = -2; x =  2