100 POINTS

Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com

fredd [130]

$f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}$

$f$ has critical points where the derivative is 0:

$2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1$

The second derivative is

$f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}$

and $f''(-1)=\frac2e>0$ , which indicates a local minimum at $x=-1$ with a value of $f(-1)=\frac1e$ .

At the endpoints of [-2, 2], we have $f(-2)=1$ and $f(2)=e^8$ , so that $f$ has an absolute minimum of $\frac1e$ and an absolute maximum of $e^8$ on [-2, 2].

So we have

$\dfrac1e\le f(x)\le e^8$

$\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx$

$\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}$

5 0

3 years ago

What is the volume of this cone?

Romashka-Z-Leto [24]

Answer:

3537.73 cubic in

Step-by-step explanation:

Given,

Diameter of the cone = 26 in

Radius of the cone = 26in ÷ 2 = 13 in

Height of the cone = 20 in

Therefore, volume of the cone

$= \frac{1}{3} \pi {r}^{2} h$

$= \frac{1}{3} \times 3.14 \times 13 \times 13 \times 20$

$= 3,537.7333333333 \: {in}^{3}$

Approx. 3537 . 73 cubic in (Ans)

8 0

3 years ago

Read 2 more answers

I have <img src="https://tex.z-dn.net/?f=f%28t%29%3D%20%20%5Cfrac%7B%20t%5E%7B2%7D%20%7D%7B9%7D%20" id="TexFormula1" title="f(t)

skad [1K]

$\bf \begin{cases}
f(t)=\cfrac{t^2}{9}\\\\
g(t)=e^{\cfrac{}{}\frac{-ln(t)}{4}}\\\\
k(t)=f(t)+c\cdot g(t)
\end{cases}\implies k(t)=\cfrac{t^2}{9}+C\cdot e^{\cfrac{}{}\frac{-ln(t)}{4}}
\\\\\\
k(1)=4\implies 
\begin{cases}
k=4\\
t=1
\end{cases}\implies 4=\cfrac{1^2}{9}+C\cdot e^{\cfrac{}{}\frac{-ln(1)}{4}}
\\\\\\
4=\cfrac{1}{9}+C\cdot e^{\frac{0}{4}}\implies 4=\cfrac{1}{9}+C\cdot 1\implies 4-\cfrac{1}{9}=C
\\\\\\
\cfrac{35}{9}=C$