Question

I have calculus problems that I need help with.

Answer 1

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

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

Evaluate this at $x=3$ and solve for $n$.

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

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

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

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

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 $x$ from each interval.

$x\in(-\infty,0) \implies x = -1 \implies f''(-1) = 32e^2 > 0$

$x\in(0,1) \implies x = \dfrac12 \implies f''\left(\dfrac12\right) = \dfrac5{43} > 0$

$x\in(1,3) \implies x = 2 \implies f''(2) = -\dfrac{16}{e^4} < 0$

$x\in(3,\infty) \implies x = 4 \implies f''(4) = \dfrac{192}{e^8} > 0$

The sign of $f''$ changes to either side of $x=1$ and $x=3$, but not $x=0$. This means only $\boxed{x=1}$ and $\boxed{x=3}$ are inflection points.

Answer 2

Answer:

Hello,

Step-by-step explanation:

A):

Max if x=3

$y=x^n*exp(-2x),\ n > 2\\\\f'(x)=n*x^{n-1}*exp(-2x)+x^n*exp(-2x)*(-2)\\=exp(-2x)*x^{n-1}*(n-2x)\\f'(x)=0\ \Longrightarrow\ x=\dfrac{n}{2} \since\ x=3\ \Longrightarrow\ n=6\\extremum\ is \ for\ n=6\\f"(x)=n(n-1)*x^{n-2}*exp(-2x)+\\n*x^{n-1}*exp(-2x)*(-2)\\+(-2)*n*x^{n-1}*exp(-2x}+4x^n*exp(-2x)\\\\...\\f"(x)=x^{n-2}*exp(-2x)*(4x^2-4*6*x+6*5)\\\\f"(x)=x^{n-2}*exp(-2x)*(x-\dfrac{6-\sqrt{6} }{2} )*(x-\dfrac{6+\sqrt{6} }{2} )\\\\f"(3)=3^4*exp(-6)*(4*3^2-4*6*3+6*5) =-1,2048... < 0\\So\ a\ maximum.$

B)

for n=4:

$f"(x)=x^2*exp(-2x)*(4x^2-16x+12)\\=4*x^2*exp(-2x)*x-3)(x-1)\\\\3\ inflexion\ points: \ x=0,\ x=1,\ x=3$