Answer:
Step-by-step explanation:
Given the limit of a function expressed as 
, to evaluate the following steps must be carried out.
Step 1: substitute x = 0 into the function
Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function
![= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20sin%28x%29-tan%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%28x%5E3%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7Bcos%28x%29-sec%5E2%28x%29%7D%7B3x%5E2%7D%5C%5C)
Step 3: substitute x = 0 into the resulting function
Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2
![= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20cos%28x%29-sec%5E2%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%283x%5E2%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B-sin%28x%29-2sec%5E2%28x%29tan%28x%29%7D%7B6x%7D%5C%5C)
Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4
![= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20-sin%28x%29-2sec%5E2%28x%29tan%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%286x%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5B%20-cos%28x%29-2%28sec%5E2%28x%29sec%5E2%28x%29%2B2sec%5E2%28x%29tan%28x%29tan%28x%29%5D%7D%7B6%7D%5C%5C%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5B%20-cos%28x%29-2%28sec%5E4%28x%29%2B2sec%5E2%28x%29tan%5E2%28x%29%5D%7D%7B6%7D%5C%5C)
Step 7: substitute x = 0 into the resulting function in step 6
![= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}](https://tex.z-dn.net/?f=%3D%20%20%5Cdfrac%7B%5B%20-cos%280%29-2%28sec%5E4%280%29%2B2sec%5E2%280%29tan%5E2%280%29%5D%7D%7B6%7D%5C%5C%5C%5C%3D%20%5Cdfrac%7B-1-2%280%29%7D%7B6%7D%20%5C%5C%3D%20%5Cdfrac%7B-1%7D%7B6%7D)
<em>Hence the limit of the function </em>
.
9514 1404 393
Answer:
B, D
Step-by-step explanation:
The cubic function is plotted in the attachment. It is a curve found in quadrants I and III, so is not symmetric about the y-axis.
Its domain and range are all real numbers, and it passes through the origin.
<span>You have:
- The diameter of the cylinder is 12 inches and its height is 14 inches.
-The height of the cone is 6 inches.
So, you must apply the formula for calculate the volume of the cylinder a the formula for calculate the volume of a cone.
V1=</span>πr²h
<span>
V1 is the volume of the cylinder.
r is the radius.
h is the height (h=14 inches)
The problem gives you the diameter, but you need the radius, so you have:
r=D/2
r=12 inches/2
r=6 inches
When you substitute the values into the formula, you obtain:
V1==</span>πr²h
V1=(3.14)(6 inches)²(14 inches)
V1=1582.56 inches³<span>
The volume of the cone is:
V2=(</span>πr²h)/3
<span>
V2 is the volume of the cone.
r is the radius (r=6 inches)
h is the height of the cone (h=6 inches).
Then, you have:
</span>
V2=(πr²h)/3
V2=(3.14)(6 inches)²(6 inches)/3
V2=226.08 inches³
<span>
Therefore, </span>the volume of the cake<span> (Vt) is:
Vt=V1+V2
Vt=</span>1582.56 inches³+226.08 inches³
<span> Vt=1808.6 inches</span>³
Answer:
(-2,-1); (-2,-4); (0,-1); (0,-3); (-1,-4); (-1,-3)
Step-by-step explanation:
Just reflect it starting at x=-2
