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Question 1008194: Find the absolute extrema of the function f(x)=1/2cos2x+(√3)sinx on the interval [0,π].
Find the slope of the line tangent to the graph of y^3-x^2y+4x=7 at the point (-2,3).
Please.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! # 1
First apply the derivative
f(x) = (1/2)*cos(2x) + sqrt(3)*sin(x)
f ' (x) = (1/2)*2*(-sin(2x)) + sqrt(3)*cos(x)
f ' (x) = -sin(2x) + sqrt(3)*cos(x)
Now solve f ' (x) = 0
f ' (x) = -sin(2x) + sqrt(3)*cos(x)
0 = -sin(2x) + sqrt(3)*cos(x)
-sin(2x) + sqrt(3)*cos(x) = 0
-2*sin(x)*cos(x) + sqrt(3)*cos(x) = 0
cos(x)*(-2*sin(x) + sqrt(3)) = 0
cos(x) = 0 or -2*sin(x) + sqrt(3) = 0
cos(x) = 0 or sin(x) = sqrt(3)/2
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If cos(x) = 0, then x = pi/2
If sin(x) = sqrt(3)/2, then x = pi/3 or x = 2pi/3
The critical values are: pi/2, pi/3, 2pi/3
Keep in mind that we're restricted to the interval [0,pi]
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Now plug each critical point into f(x). Also, plug each endpoint (0 and pi) into f(x) as well. The smallest output will be the absolute minimum. The largest output will be the absolute minimum.
This is the list of values we'll plug in:
0, pi/3, pi/2, 2pi/3, pi
Plug in x = 0 into f(x) to get...
f(x) = (1/2)*cos(2x) + sqrt(3)*sin(x)
f(0) = (1/2)*cos(2*0) + sqrt(3)*sin(0)
f(0) = 0.5
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Plug in x = pi/3 into f(x) to get...
f(x) = (1/2)*cos(2x) + sqrt(3)*sin(x)
f(pi/3) = (1/2)*cos(2*pi/3) + sqrt(3)*sin(pi/3)
f(pi/3) = 1.25
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Plug in x = pi/2 into f(x) to get...
f(x) = (1/2)*cos(2x) + sqrt(3)*sin(x)
f(pi/2) = (1/2)*cos(2*pi/2) + sqrt(3)*sin(pi/2)
f(pi/2) = 1.232051
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Plug in x = 2pi/3 into f(x) to get...
f(x) = (1/2)*cos(2x) + sqrt(3)*sin(x)
f(2pi/3) = (1/2)*cos(2*2pi/3) + sqrt(3)*sin(2pi/3)
f(2pi/3) = 1.25
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Plug in x = pi into f(x) to get...
f(x) = (1/2)*cos(2x) + sqrt(3)*sin(x)
f(pi) = (1/2)*cos(2*pi) + sqrt(3)*sin(pi)
f(pi) = 0.5
The smallest output was 0.5 (happens when either x = 0 or x = pi)
The largest output was 1.25 (happens when either x = pi/3 or x = 2pi/3)
Absolute Min: 0.5
Absolute Max: 1.25
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# 2
Find the slope of the line tangent to the graph of y^3-x^2y+4x=7 at the point (-2,3).
Implicit Differentiation
y^3-x^2y+4x=7
d/dx[y^3-x^2y+4x]=d/dx[7]
3y^2*dy/dx-2xy-x^2*dy/dx+4 = 0
3y^2*dy/dx-x^2*dy/dx = 2xy-4
dy/dx(3y^2-x^2) = 2xy-4
dy/dx = (2xy-4)/(3y^2-x^2)
Now plug in (x,y) = (-2,3)
dy/dx = (2xy-4)/(3y^2-x^2)
dy/dx = (2*(-2)*(3)-4)/(3(3)^2-(-2)^2)
dy/dx = -16/23
The slope of the tangent line at (-2,3) is -16/23
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