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BigorU [14]
3 years ago
12

Help please will vote you as brainliest

Mathematics
2 answers:
katrin2010 [14]3 years ago
7 0
Question One: A
Question Two: C
Question  Three: B

I'm certain that One and Two are right, I'm pretty sure Three is right.  I hope this helps.

denis23 [38]3 years ago
4 0
First answer is A
Second answer is C

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

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f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1

The gradient at x = 0 is given by evaluating f'(0).

f'(0) = 3(0)^2 - 1 = -1

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.

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

y = -x + c

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).

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So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}
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Step-by-step explanation:

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