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Let f(x)=x^3-x-1

a. Find the equation of the tangent line to the graph of the function at x = 0.


b. What does the derivative of the function at x = 0 tell you about the direction at that point? Is it increasing, decreasing, or neither?

Sagot :

superkoopasirf
[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]
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