Resuelve las siguientes ecuaciones
1 answer:
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Answer: D
Step-by-step explanation:
To find the difference, we want to make sure both denominators are equal.
Now that the denominators are equal, we can distribute and multiply out.
With the fractions multiplied out, we can actually put it into one large fraction.
Let's subtract the top expression.
This may seem like our final answer, but we can actually factor out an 3x in the numerator and denominator.
With the 3x factored out, they cancel out because 3x/3x=1.
Our final answer is:
The picture below is the answer to your question. I solved it myself when I was doing my homework.
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.