Given f(t) =t^2 − t − ln(t)

Question

a) Find the intervals on which f is increasing or decreasing,

b) Find the local maximum and minimum values of f

c) Find the interval of concavity and inflection points

Answer 1

Graph the equation!!!

You will see that this is a modified parabola which goes to ∞ at 0 and as t increases and has a minimum at t=1.

Answer 2

a) Find the intervals on which f is increasing or decreasing,

Find this from the 1st derivative---which will be + where f is increasing, and - where it is decreasing

b) Find the local maximum and minimum values of f

Maximums and minimums correspond to the second derivative going to zero.

c) Find the interval of concavity and inflection points

For this, use the second derivative:

https://www.dummies.com/education/math/calculus/how-to-locate-intervals-of-concavity-and-inflection-points/

Answer 3

First, you need to look at the domain of f(t) = t² - t - ln(t). Since ln(t) is not defined for t ≤ 0, the domain of this function is (0,∞).

a) You can determine the intervals where an equation is increasing or decreasing by looking at where the derivative – f'(t) – is greater than or less than 0. Values of t for f'(t) > 0 are where the function is increasing and values of t for f'(t) < 0 are where the function is decreasing.

f'(t) > 0

2t - 1 - (1/t) > 0

2t² - t - 1 > 0

(2t + 1)(t - 1) > 0

2t + 1 > 0 & t - 1 > 0

t > -1/2 & t > 1

t = -1/2 is not in domain, therefore the function increases for t > 1 or (1, ∞).

f'(t) < 0

2t² - t - 1 < 0

(2t + 1)(t - 1) < 0

t = -1/2 not in domain, so the function decreases over the interval of 0 < t < 1 or (0, 1).

b) Local min and max points are determined by values of t for f'(t) = 0. Similar to part a, t = -1/2 calculates as a possible local min or max for the original function. However, since t = -1/2 is not in the domain, it cannot be a solution. Therefore, t = 1 is the only solution for a local min/max.

f'(t) = 0

2t - 1 - (1/t) = 0

2t² - t - 1 = 0

(2t + 1)(t - 1) = 0

2t + 1 = 0 & t - 1 = 0

t = -1/2 & t = 1

t = -1/2 not in domain; t = 1

c) Inflection points and concavity are found using the second derivative – f''(t). Inflection points are determined by f''(t) = 0.

f''(t) = 0

2 + (1/t²) = 0

1 = -2t²

t² = -1/2

Since a square can never equal a negative number, there are no inflection points.

Concavity is found by answering the question: is f''(t) > 0 or f''(t) < 0?

If you try to solve f''(t) < 0, you end up with an equation like this.

2 + (1/t²) < 0

2t² < -1

t² < -1/2

t² < -1/2 cannot exist therefore the graph is never concave down. Conversely, t² will always be greater than -1/2 by virtue of squares always being positive. So, f''(t) > 0 and the graph of the original equation is concave up.