1. To find the gradient of the curve at x=1, we need to find the derivative of f(x) and evaluate it at x=1.
f'(x) = 3x^2 - 12x - 15
f'(1) = 3(1)^2 - 12(1) - 15 = -24
Therefore, the gradient of the curve at x=1 is -24.
2. To find the maximum and minimum points, we need to find the stationary points first by setting f'(x) equal to 0.
f'(x) = 3x^2 - 12x - 15 = 0
Simplifying, we get:
x^2 - 4x - 5 = 0
Using the quadratic formula, we get:
x = (-(-4) +/- sqrt((-4)^2 - 4(1)(-5))) / (2(1))
x = 5 or x = -1
Now we need to determine whether these stationary points are maximum or minimum points. We can use the second derivative test for this.
f''(x) = 6x - 12
At x=5:
f''(5) = 6(5) - 12 = 18
Since the second derivative is positive, f(x) has a minimum point at x=5.
At x=-1:
f''(-1) = 6(-1) - 12 = -18
Since the second derivative is negative, f(x) has a maximum point at x=-1.
Therefore, the maximum point is (-1, f(-1)) = (-1, 13) and the minimum point is (5, f(5)) = (5, -91).
A curve is defined by f(x)=x^3-6x^2-15x-1. Find the
1.gradient of the curve at x=1
2.maximum and minimum points
1 answer
1 answer