The answer to this is 14/16
Both equations have
as the left hand side.
Then, in a situation like
![y=a,\quad y=b](https://tex.z-dn.net/?f=y%3Da%2C%5Cquad%20y%3Db)
we can deduce
, since they both equal ![y](https://tex.z-dn.net/?f=y)
So, we can set up the equation
![4x^2-6x+4=x+1 \iff 4x^2-7x+3 = 0](https://tex.z-dn.net/?f=%204x%5E2-6x%2B4%3Dx%2B1%20%5Ciff%204x%5E2-7x%2B3%20%3D%200)
The solutions are
![x=\dfrac{3}{4},\quad x=1](https://tex.z-dn.net/?f=%20x%3D%5Cdfrac%7B3%7D%7B4%7D%2C%5Cquad%20x%3D1%20)
From the second equation we know that y is one more than x, so we have
![x=\dfrac{3}{4} \implies y = \dfrac{7}{4},\quad x=1 \implies y=2](https://tex.z-dn.net/?f=%20x%3D%5Cdfrac%7B3%7D%7B4%7D%20%5Cimplies%20y%20%3D%20%5Cdfrac%7B7%7D%7B4%7D%2C%5Cquad%20x%3D1%20%5Cimplies%20y%3D2%20)
Answer:
0.95
Step-by-step explanation:
The answer is 3.92.
Hope this helps!
9514 1404 393
Answer:
x = 1
Step-by-step explanation:
The graph of f(x) has a relative minimum where its first derivative (f'(x)) is zero and its second derivative (f''(x)) is positive. (Function f(x) is concave upward when f''(x) > 0.) Since the second derivative is the slope of the first derivative, this means the local minimum will be at the x-value where f'(x) = 0 and is crossing the x-axis from below.
The graph of f'(x) shows a local minimum at x = 1.
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Desmos does not allow us to define f'(x) directly, so we have called it f₁(x). The function f(x) is the integral of that—shown as the dashed orange line. We wanted you to see the f(x) curve so you could see it has a local minimum at x=1.