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tan2 a then find the minimum value of Vx2 Ifαε 3. 0, X 2 tan2 a then find the minimum value of Vx2 Ifαε 3. 0, X 2
tan2 IX (a) lim x+0 sinx tan2 IX (a) lim x+0 sinx
solve the equation on the interval [0,360) 3) 4 tan2 + 7 tan 0 - 2 = 0 4) 3 sin2 x + sin x = 0
sin x aln3' nena) xln tan1 (x - Vx2 +1) Q1/Find the Derivatives: 1) y 2) = y ln3 c logs x2 b lnx 3) y ae* + sin x aln3' nena) xln tan1 (x - Vx2 +1) Q1/Find the Derivatives: 1) y 2) = y ln3 c logs x2 b lnx 3) y ae* +
at the point (2, -3). Give an 4. Find the minimum value of the directional derivative of the function f(x, y)- exact answer. 2. at the point (2, -3). Give an 4. Find the minimum value of the directional derivative of the function f(x, y)- exact answer. 2.
Find the absolute maximum value and absolute minimum value of the function f ( x ) = 4sin^3 x + 3cos^2 x on [ 0,π ] .
10) Integrate f(x, y) = sin (Vx2 + y2) over the region 0 < x2 + y2 = 16
b. 2. (12 pts.) Find the exact value. sin(1659) c. sin(52.5°) sin(7.5°) tan(159) 1 - tan2(15°) a.
1. 2. Find the maximum and minimum value of f(x,y) = x² ty? - xy +1 on the triangles region R with vertices (0,0), (2,), (0, 2)
Solve the equation. (Use n as an integer constant, where x = [0, a) when n = 0.) (3 tan? x - 1)(tan? x – 3) = 0 3 tan2 x - 1 x= (smaller value for n = 0) 3 tan? x – 1 x = (larger value for n = 0) X tan? x - 3 = (smaller value for n = 0) X tan2 x – 3 x (larger value for n = 0)
how you solve this? what are the steps? 2. Find the x-value representing the absolute minimum value of... f(x) = x3 - 9x2 + 24x; (0,5), and find the y-value of the minimum value.