Answer :
[tex]f(x) = x^3 - x - 1[/tex]
To find the gradient of the tangent, we must first differentiate the function.
[tex]f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1 [/tex]
The gradient at x = 0 is given by evaluating f'(0).
[tex]f'(0) = 3(0)^2 - 1 = -1[/tex]
The derivative of the function at this point is negative, which tells us the function is decreasing at that point.
The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so
[tex]y = -x + c[/tex]
Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).
[tex]f(0) = (0)^3 - (0) - 1 = -1[/tex]
So the point (0, -1) lies on the tangent. Substituting into the tangent equation:
[tex]y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}[/tex]
To find the gradient of the tangent, we must first differentiate the function.
[tex]f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1 [/tex]
The gradient at x = 0 is given by evaluating f'(0).
[tex]f'(0) = 3(0)^2 - 1 = -1[/tex]
The derivative of the function at this point is negative, which tells us the function is decreasing at that point.
The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so
[tex]y = -x + c[/tex]
Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).
[tex]f(0) = (0)^3 - (0) - 1 = -1[/tex]
So the point (0, -1) lies on the tangent. Substituting into the tangent equation:
[tex]y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}[/tex]
