Derivatives and Shape of Graph

Question

Consider the function q(t)=e^t/(3+e^t)

interval(s) on which q is increasing:

interval(s) on which q is decreasing:

point(s) at which q achieves a local maximum:

point(s) at which q achieves a local minimum:

interval(s) on which q concaves up:

interval(s) on which q concaves down:

Answer 1

q(t) = eᵗ/(3 + eᵗ)

horizontal asymptotes

q(t)→ 0 as t → –∞

q(t)→ 1 as t → ∞

first derivative

q'(t) = [(3 + eᵗ)eᵗ – e²ᵗ]/(3 + eᵗ)²

= (3eᵗ + e²ᵗ – e²ᵗ)/(3 + eᵗ)²

= 3eᵗ/(3 + eᵗ)²

q'(t) > 0 for all t so q(t) is always increasing

second derivative

q''(t) = [3(3 + eᵗ)²eᵗ – 2(3 + eᵗ)3e²ᵗ]/(3 + eᵗ)⁴

= [3(9eᵗ + 6e²ᵗ + e³ᵗ) – 6(3e²ᵗ + e³ᵗ)]/(3 + eᵗ)⁴ =

[27eᵗ + (18 – 18)e²ᵗ + (3 – 6)e³ᵗ]/(3 + eᵗ)⁴ =

(27eᵗ – 3e³ᵗ)/(3 + eᵗ)⁴ = 3eᵗ(9 – e²ᵗ)/(3 + eᵗ)⁴

q''(t) = 0 when e²ᵗ = 9 which means eᵗ = 3

and t = ln(3)

So q(t) has an inflection point at t = ln(3)

curvature (concavity)

q''(t) > 0 when t < ln(3) which means

q(t) is concave up when t < ln(3)

q''(t) < 0 when t > ln(3) which means

q(t) is concave down when t > ln(3)

Answers

interval(s) on which q is increasing:

(–∞,∞)

interval(s) on which q is decreasing:

null (q is never decreasing)

point(s) at which q achieves a local maximum:

none

point(s) at which q achieves a local minimum:

none

interval(s) on which q is concave up:

(–∞, ln3)

interval(s) on which q is concave down:

(ln3, ∞)