Question

Concavity Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$f(x)=x^{4}-2 x^{3}+1$$

Short answer

Question: Determine the intervals of concavity and inflection points for the function $$f(x) = x^{4} - 2x^{3} + 1$$. Answer: The function is concave up on the intervals $$(-\infty, 0)$$ and $$(1, \infty)$$, and concave down on the interval $$(0, 1)$$. The inflection points are $$(0, 1)$$ and $$(1, 0)$$.

Step-by-step solution

Find the first derivative of the function

Differentiate the given function $$f(x)=x^{4}-2 x^{3}+1$$ with respect to $$x$$: $$f'(x) = 4x^{3} - 6x^{2}$$

Find the second derivative of the function

Differentiate the first derivative $$f'(x) = 4x^{3} - 6x^{2}$$ with respect to $$x$$: $$f''(x) = 12x^{2} - 12x$$

Factor the second derivative

Factor $$f''(x) = 12x^{2} - 12x$$ to find critical points: $$f''(x)= 12x(x - 1)$$

Solve for critical points

Find the critical points by solving $$f''(x) = 0$$: $$12x(x - 1) = 0$$ This gives us two critical points, $$x=0$$ and $$x=1$$.

Determine the intervals of concavity and inflection points

Create a sign chart for the second derivative to find the intervals of concavity: - $$(-\infty, 0)$$: Choose a number less than 0, say $$x=-1$$. Then, $$f''(-1) = 12(-1)(-2) > 0$$, so the function is concave up on this interval. - $$(0, 1)$$: Choose a number between 0 and 1, say $$x=0.5$$. Then, $$f''(0.5) = 12(0.5)(-0.5) < 0$$, so the function is concave down on this interval. - $$(1, \infty)$$: Choose a number greater than 1, say $$x=2$$. Then, $$f''(2) = 12(2)(1) > 0$$, so the function is concave up on this interval. Thus, the function is concave up on the intervals \((-\infty, 0)\) and \((1, \infty)\), and concave down on the interval \((0, 1)\). The inflection points are found at the critical points $$x=0$$ and $$x=1$$, and their corresponding $$y$$-coordinates are: $$f(0) = (0)^{4} - 2(0)^{3} + 1 = 1$$ $$f(1) = (1)^{4} - 2(1)^{3} + 1 = 0$$ So, the inflection points are $$(0, 1)$$ and $$(1, 0)$$.

Key concepts

Inflection Points

Inflection points occur where the concavity of a function changes. These are the points where the curve changes direction from concave up (bowl-shaped) to concave down (hill-shaped) or vice versa. To find inflection points mathematically, we look for points where the second derivative of a function changes its sign.

In the given problem, the second derivative of the function is factored as \(f''(x) = 12x(x-1)\). The critical points obtained from setting this to zero are \(x = 0\) and \(x = 1\). These are potential inflection points.

After determining the intervals of concavity, we validate that the change in sign of \(f''(x)\) at \(x = 0\) and \(x = 1\) confirms their status as inflection points. Thus, the function's curve changes its concavity at these points. Evaluated y-values at these x-coordinates give us the inflection points \((0, 1)\) and \((1, 0)\).

Intervals of Concavity

Intervals of concavity tell us where a function is curving upwards or downwards. A function is **concave up** when the graph bends upwards like a cup, and **concave down** when it curves downwards like an arch. The second derivative \(f''(x)\) helps us find these intervals.

By analyzing the sign of \(f''(x)\), we can find where the function is concave up or concave down. Choose test points in intervals created by critical points to determine this:

• For \((-\infty, 0)\), using the test point \(x = -1\), \(f''(-1) > 0\), indicating concave up.
• For \((0, 1)\), using the test point \(x = 0.5\), \(f''(0.5) < 0\), indicating concave down.
• For \((1, \infty)\), using the test point \(x = 2\), \(f''(2) > 0\), indicating concave up.

This gives us the result that the function is concave up on \((-\infty, 0)\) and \((1, \infty)\), and concave down on \((0, 1)\).

Second Derivative

The second derivative helps us understand the curvature of a function's graph, essentially informing us about concavity.

The second derivative is derived from differentiating the first derivative. For the function \(f(x) = x^4 - 2x^3 + 1\), the first derivative is \(f'(x) = 4x^3 - 6x^2\). Differentiating this once more gives the second derivative \(f''(x) = 12x^2 - 12x\).

Factoring the second derivative to \(f''(x) = 12x(x - 1)\) makes it easier to find critical points where \(f''(x) = 0\). These critical points help in further analysis of the function's concavity and finding inflection points.

Critical Points

Critical points of \(f''(x)\) are crucial in identifying where a function may change its curviness, or concavity. They are points on the graph where the slope of the tangent is zero or undefined for \(f'(x)\). However, when it comes to concavity, we specifically look where the second derivative is zero.

In the exercise, to find the function's critical points, you solve \(f''(x) = 0\). With \(f''(x) = 12x(x - 1)\), setting it to zero gives \(x = 0\) and \(x = 1\) as critical points.

These critical points divide the x-axis into intervals to test for concavity. While these points are potential inflection points, remember: make sure to confirm that they indeed change the sign of the second derivative before concluding.