Given that 16 ounces equals 1 pound, how many ounces are in 12.4 pounds?
2 answers:
For this case we must find the number of ounces that there are in 12.4 pounds, if we have as data that 16 ounces represent 1 pound, then we can make a rule of three of the form:
16 ounces -----------> 1 pound
x ----------------------> 12.4 pounds
Where "x" represents the number of ounces in 12.4 pounds. So, we have:
So, in 12.4 pounds there are 198.4 ounces
Answer:
198.4 ounces
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A)
why is 0? well, the bounds are the same.
b)
let's use the second fundamental theorem of calculus, where F'(x) = dF/du * du/dx
![\bf \displaystyle F(x)=\int\limits_{4}^{u}\cfrac{2t-1}{t+2}\cdot dt\qquad F'(x)=\cfrac{dF}{du}\cdot \cfrac{du}{dx} \\\\\\ \displaystyle\cfrac{d}{du}\left[ \int\limits_{4}^{u}\cfrac{2t-1}{t+2}\cdot dt \right]\cdot \cfrac{du}{dx}\implies \cfrac{2u-1}{u+2}\cdot \cfrac{1}{2\sqrt{x}} \\\\\\ \cfrac{2\sqrt{x}}{\sqrt{x}+2}\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{2\sqrt{x}-1}{2x+4\sqrt{x}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cdisplaystyle%20F%28x%29%3D%5Cint%5Climits_%7B4%7D%5E%7Bu%7D%5Ccfrac%7B2t-1%7D%7Bt%2B2%7D%5Ccdot%20dt%5Cqquad%20F%27%28x%29%3D%5Ccfrac%7BdF%7D%7Bdu%7D%5Ccdot%20%5Ccfrac%7Bdu%7D%7Bdx%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle%5Ccfrac%7Bd%7D%7Bdu%7D%5Cleft%5B%20%5Cint%5Climits_%7B4%7D%5E%7Bu%7D%5Ccfrac%7B2t-1%7D%7Bt%2B2%7D%5Ccdot%20dt%20%5Cright%5D%5Ccdot%20%5Ccfrac%7Bdu%7D%7Bdx%7D%5Cimplies%20%5Ccfrac%7B2u-1%7D%7Bu%2B2%7D%5Ccdot%20%5Ccfrac%7B1%7D%7B2%5Csqrt%7Bx%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2%5Csqrt%7Bx%7D%7D%7B%5Csqrt%7Bx%7D%2B2%7D%5Ccdot%20%5Ccfrac%7B1%7D%7B2%5Csqrt%7Bx%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%7Bx%7D-1%7D%7B2x%2B4%5Csqrt%7Bx%7D%7D)
c)
we know x = 16, we also know from section a) that at that point f(x) = y = 0, so the point is at (16, 0), using section b) let's get the slope,
d)
now, we can get critical points from zeroing out the derivative, we also get critical points from zeroing out the denominator, however, the ones from the denominator are points where the function is not differentiable, namely, is not a smooth curve, is a sharp jump, a cusp, or a spike, and therefore those points are usually asymptotic, however, they're valid critical points, let's check both,
now, doing a first-derivative test on those regions, we get the values as in the picture below.
so, you can see where is increasing and decreasing.
why is 0? well, the bounds are the same.
b)
let's use the second fundamental theorem of calculus, where F'(x) = dF/du * du/dx
c)
we know x = 16, we also know from section a) that at that point f(x) = y = 0, so the point is at (16, 0), using section b) let's get the slope,
d)
now, we can get critical points from zeroing out the derivative, we also get critical points from zeroing out the denominator, however, the ones from the denominator are points where the function is not differentiable, namely, is not a smooth curve, is a sharp jump, a cusp, or a spike, and therefore those points are usually asymptotic, however, they're valid critical points, let's check both,
now, doing a first-derivative test on those regions, we get the values as in the picture below.
so, you can see where is increasing and decreasing.