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Nutka1998 [239]
3 years ago
7

Sara is completing the square to find the maximum or minimum value of the function f(x) = (2 - x) (5 + x). What is the first ste

p that Sara must take? Does the function have a maximum or minimum value? What is that value?
A) multiply the binomials; maximum value;
49
4
B) multiply the binomials; minimum value;
49
4
C) set each factor equal to zero and solve for x; minimum value;
−2
5
D) set each factor equal to zero and solve for x; maximum value;
−7
2
Mathematics
1 answer:
DanielleElmas [232]3 years ago
4 0
Definitely multiply out the given factors:

f(x) = 10 + 2x - 5x - x^2, or  f(x) = -x^2 - 3x + 10
Find the derivative:  f '(x) = -2x - 3
Set the deriv. = to 0 and solve for x:  -2x = 3, and x = -3/2
This x = -3/2 is the x-coordinate of the max value.  The y-coord. is

f(-3/2) = (-3/2)^2 - 3(-3/2) + 10 = 21.25

I realize that this result does not agree with any of the four possible answers.  Please ensure that y ou have copied down this problem completely and correctly.

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Consider the surface defined by the equation x3 + y3 + z3 = 3xyz. Use implicit differ- ∂z entiation to find ∂x. (Note z is not a
Aloiza [94]

Answer:    

   \frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}

Step-by-step explanation:

Given equation is x^{3}+y^{3}+z^{3}=3xyz ………………(1)

we use derivative formula \frac{d}{dx}(x^{n})  = n x^{n-1}

\frac{d}{dx}(x^{3)}  = 3 x^{2}

\frac{d}{dx}(z^{3)}  = 3 z^{2}

And also apply 'u v' formula

\frac{d}{dx}(uv})  = u\frac{d}{dx}(v)+v\frac{d}{dx}(u)

Differentiating  equation (1) partially with respective to 'x' , we treated 'y' as constant.

(3x^{2} +0+3z^{2}\frac{∂z}{∂x}) =3y(x\frac{∂z}{∂x}+z(1))   ( here y treated as constant so the derivative of constant function is zero in addition but in multiplication the constant is keep as like 'y').

on simplification , we get

(3x^{2} +0+3z^{2}\frac{∂z}{∂x}) =(3yx\frac{∂z}{∂x}+3yz)

again simplification, we get

3z^{2}\frac{∂z}{∂x}- 3yx\frac{∂z}{∂x}=3yz-3x^{2}

taking common '\frac{∂z}{∂x} on left on side , we get

(3z^{2}- 3yx)\frac{∂z}{∂x}=3yz-3x^{2}

dividing '(3z^{2}- 3yx) on both sides, we get

\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}

<u>Final answer</u>:-

\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}

6 0
3 years ago
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