Question

Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$g(t)=3 t^{5}-30 t^{4}+80 t^{3}+100$$

Short answer

Question: Determine the inflection points and intervals of concavity for the function $$g(t) = 3t^{5} - 30t^4 + 80t^3 + 100$$. Answer: The function has two inflection points: (2, -44) and (4, 292). It is concave up on the intervals (-∞, 0), (0, 2), and (4, ∞) and concave down on the interval (2, 4).

Step-by-step solution

Find the first derivative of g(t)

To find the first derivative of the function, use the power rule, which states that the derivative of $$x^n$$ is $$nx^{n-1}$$. So, differentiate each term in g(t) with respect to t: $$g'(t) = 15t^{4} - 120t^{3} + 240t^{2}$$

Find the second derivative of g(t)

Again, we will use the power rule for differentiation. Differentiate each term in g'(t) with respect to t: $$g''(t) = 60t^{3} - 360t^{2} + 480t$$

Find when g''(t) changes signs to find inflection points and intervals of concavity

To find when g''(t) changes its sign, we have to find its critical points, i.e. when $$g''(t) = 0$$. So, we will set the second derivative equal to zero and solve for t: $$0 = 60t^{3} - 360t^{2} + 480t$$ Factoring out the common factor of 60t, we get: $$0 = 60t(t^{2} - 6t + 8)$$ We can further factor the quadratic expression: $$0 = 60t(t - 4)(t - 2)$$ Thus, t = 0, t = 2, and t = 4 are critical points of the second derivative. Now, we will determine where g''(t) is positive or negative by using these critical points to create intervals. We can do this by choosing test points in each interval and check the value of g''(t): 1. Interval (-∞, 0): Test point t = -1 $$g''(-1) = 60(-1)((-1) - 4)((-1) - 2) = 60(1)(3)(5) > 0$$ So, the function is concave up in this interval. 2. Interval (0, 2): Test point t = 1 $$g''(1) = 60(1)(1 - 4)(1 - 2) = 60(1)(-3)(-1) > 0$$ The function is concave up in this interval as well. 3. Interval (2, 4): Test point t = 3 $$g''(3) = 60(3)(3 - 4)(3 - 2) = 60(3)(-1)(1) < 0$$ The function is concave down in this interval. 4. Interval (4, ∞): Test point t = 5 $$g''(5) = 60(5)(5 - 4)(5 - 2) = 60(5)(1)(3) > 0$$ The function is concave up in this interval.

Identify inflection points and intervals of concavity

Based on the intervals in step 3, the function has two inflection points: 1. t = 2: This is an inflection point since the function changes from concave up to concave down. The point is g(2) = 3(2)^{5}-30 (2)^{4}+80 (2)^{3}+100 = -44. 2. t = 4: This is an inflection point since the function changes from concave down to concave up. The point is g(4) = 3(4)^{5}-30 (4)^{4}+80 (4)^{3}+100 = 292. The function is concave up on the intervals (-∞, 0), (0, 2), and (4, ∞) and concave down on the interval (2, 4).

Key concepts

Inflection Points

Inflection points are pivotal in understanding the concavity attributes of a function. To pinpoint these points, we first need to find where the function changes its curve from concave up to concave down or vice versa. How do we do this? By analyzing the second derivative of the function. The second derivative, denoted as \( g''(t) \), provides vital clues to these changes.

Let's say, in our function \( g(t) \), we calculate the second derivative and set it equal to zero: \( 0 = 60t(t - 4)(t - 2) \). Solving for \( t \), we find critical points that may be candidates for inflection points.

So, inflection points occur at \( t = 2 \) and \( t = 4 \) because the function changes its concavity there.

• At \( t = 2 \), \( g(t) \) shifts from being concave up to concave down.
• At \( t = 4 \), \( g(t) \) then transitions from concave down to concave up.

This precise transition marks the presence of an inflection point.

Critical Points

Critical points are a fundamental concept when analyzing functions. These points occur where the first derivative of the function equals zero or does not exist.

However, when exploring inflection points, we concern ourselves with the second derivative, \( g''(t) \), and specifically seek where \( g''(t) = 0 \) or changes its sign. In the provided solution, by setting \( 60t(t - 4)(t - 2) = 0 \), we acquired the critical values \( t = 0, 2, 4 \). Each of these signals a potential inflection point and helps in categorizing the nature of the function’s curve over certain intervals.

These critical points segment the number line into intervals which we test further to affirm the function's concavity nature at given spans.

Second Derivative

The second derivative is an extension of differential calculus that informs us not only about the curve's slope but also about its concavity - whether it's curving upwards like a smile or downwards like a frown.

In our example, \( g''(t) = 60t^{3} - 360t^{2} + 480t \), describes how the rate of change itself is changing. The second derivative is crucial because:

• A positive \( g''(t) \) suggests the function is concave up.
• A negative \( g''(t) \) indicates the function is concave down.
• When \( g''(t) = 0 \), it reveals candidates for inflection points, where concavity switches.

By examining test points in each interval defined by the critical points, we determine whether \( g(t) \) is concave up or down in those stretches. This comprehensive approach leads us to a deeper understanding of the function's behavior over its entire domain.