I have calculus problems that I need help with.

Question

asked Feb 9, 2023 157k views

2 Answers

AlexITC · Answer 1 · 2023-02-14T14:43:06+0000

a. Note that
f(x)=x^ne^(-2x) is continuous for all
. If
f(x) attains a maximum at
x=3 , then
f'(3) = 0 . Compute the derivative of
.

f'(x) = nx^(n-1) e^(-2x) - 2x^n e^(-2x)

Evaluate this at
x=3 and solve for
.

$n\cdot3^(n-1) e^(-6) - 2\cdot3^n e^(-6) = 0$

$n\cdot3^(n-1) = 2\cdot3^n$

$\frac n2 = (3^n)/(3^(n-1))$

$\frac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of
, we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^(-2x) \\\\ \implies f'(x) = 6x^5 e^(-2x) - 2x^6 e^(-2x) \\\\ \implies f''(x) = 30x^4 e^(-2x) - 24x^5 e^(-2x) + 4x^6 e^(-2x) \\\\ \implies f''(3) = -(486)/(e^6) < 0$

The second derivative at
x=3 is negative, which indicate the function is concave downward, which in turn means that
f(3) is indeed a (local) maximum.

b. When
n=4 , we have derivatives

$f(x) = x^4 e^(-2x) \\\\ \implies f'(x) = 4x^3 e^(-2x) - 2x^4 e^(-2x) \\\\ \implies f''(x) = 12x^2 e^(-2x) - 16x^3e^(-2x) + 4x^4e^(-2x)$

Inflection points can occur where the second derivative vanishes.

12x^2 e^(-2x) - 16x^3 e^(-2x) + 4x^4 e^(-2x) = 0

12x^2 - 16x^3 + 4x^4 = 0

4x^2 (3 - 4x + x^2) = 0

4x^2 (x - 3) (x - 1) = 0

Then we have three possible inflection points when
x=0 ,
x=1 , or
x=3 .

To decide which are actually inflection points, check the sign of
f'' in each of the intervals
$(-\infty,0)$ ,
(0, 1) ,
(1, 3) , and
$(3,\infty)$ . It's enough to check the sign of any test value of
from each interval.