f(x)=x^3-6x+1
find the critical points, where the function is increasing or decreasing and explain the shape of the graph using the derivative and algebra.
i found the derivative to be 3x^2-6. So the critical points are -1.4 and 1.4. The derivative is decreasing until (0,6) and then it begins increasing. i don't know how to go about explaining the shape of the graph with algebra though. is it related to quadratic equations???
yes. Now, you need the second derivative, which in this cae is 6x. So, at the first "critical point,-1.4, the second derivative is negative, so at that point, so it is a maximum. At the second point 1.4, the second derivative is positive, so the graph is a relative minumum there. So, the graph curves up till it hits -1.4, then it curves down unitl it hits 1.4, then it curves up again.
To find the critical points and determine where the function is increasing or decreasing, we first need to find the derivative of the function f(x)=x^3-6x+1. You correctly calculated the derivative as f'(x)=3x^2-6.
Critical points occur when the derivative is equal to zero or undefined. In this case, since the derivative is a polynomial, it is defined for all real numbers. So we only need to find where it is equal to zero.
Setting the derivative equal to zero:
3x^2-6=0
Dividing both sides by 3:
x^2-2=0
Now, we can solve for x by factoring:
(x+√2)(x-√2)=0
Therefore, the critical points are x=-√2 and x=√2. However, it seems there was an error in your calculation. The value of √2 is approximately 1.4, not 1.7.
To determine where the function is increasing or decreasing, we can analyze the sign of the derivative on different intervals. We can create a number line and test points within each interval.
Let's consider three intervals: (-∞,-√2), (-√2,√2), and (√2,∞).
Pick a test point from each interval and substitute it into the derivative:
For the interval (-∞,-√2), we can choose x=-3:
f'(-3)=(3(-3)^2)-6=27-6=21>0
Since the derivative is positive in this interval, the function is increasing.
For the interval (-√2,√2), we can choose x=0:
f'(0)=3(0^2)-6=-6<0
Since the derivative is negative in this interval, the function is decreasing.
For the interval (√2,∞), we can choose x=3:
f'(3)=(3(3)^2)-6=27-6=21>0
Again, the derivative is positive, so the function is increasing.
Now let's discuss the shape of the graph using this information. Since the function is increasing on the interval (-∞,-√2) and again on the interval (√2,∞), it means the function is going upward as x approaches negative infinity and as x approaches positive infinity.
Since the function is decreasing on the interval (-√2,√2), the function reaches a local minimum at x=√2 and a local maximum at x=-√2. The graph changes from increasing to decreasing at these points.
Additionally, based on the sign of the second derivative (which is 6x), we can determine whether the function is concave upward or downward. In this case, since the coefficient of the second derivative is positive, the function is concave upward. This means the curve is shaped like a cup facing upwards, with the local minimum and maximum points being the bottom and top of the cup, respectively.
Algebraically, the shape of the graph can also be analyzed using the concavity test. Since the second derivative is positive for all x values, the graph is always concave upward. This concavity indicates the shape of the curve as it bends smoothly and opens upward, consistent with a cubic polynomial.