Values Essay Examples

values (if they exist) 2 4 6 8 10 12 14 1 0 1 (b) For the function find and an equation of the tangent line to the graph of the above function at In the questions c to h , find (C ) y = x3 – x2 – x5/3 = 3x2 – 2x – 5/3x (d) y=x3 ln(1+x2) You may use (e) y'= 2+cosx+ ddxsinx-sinx. ddx(2+cosx)(2+cosx)2y'= 2+cosx.cosx-sinx. (-sin⁡(x))(2+cosx)2y'= 2+cosx+cos2 x+sin2(x)(2+cosx)2cos2 x+sin2x=1sin2x=1- cos2 xy'= 2+cosx+cos2 x+1- cos2 x(2+cosx)2y'= 1+ 2cosx(2+cosx)2(f) y=sin-1(ex)siny=(ex)cosyy'=(ex)y'=excosyy=ex1-e2x(g) Find if i.e. = ddxsin3(2x)=3sin22x.cos2x.2(2) (a) Find an equation of the tangent line to the graph of y=2xcosx at 0,1 Use Geogebra or any...

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Algebra

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values would be real except when the value of x=4 hence the domain is D: allR: x≠4Question 4 Slope (m) = -3/7, y-intercept(c) =-4Y=mx+cHence, y=-37x -4Question 5 Slope(m) = -2, passes through (-1,6) hence the y intercept is Taking a point (x,y) where the line passes then Gradient is ∆y/∆x = y-6x-(-1)=-2y-6x+1=-2Hence y=-2x+4Question 6 (2x-7)(x+4)=02x(x+4)-7(x+4)=02x(x+4)=7(x+4)2x2+8x=7x+28 hence 2x2+x-28=0Using the factor method the values of X=7/2 , -4Question 7 Vertex f(x) =3x2-6x+8f(x) = a(x - h)2 + k, where (h, k) is the vertex hence f(x) =3(x-1)2+5the vertex =(1,5)Question 8 Y intercept of f(x) =(x-2)(x+1)(x-3)=(x^2-x-2)(X-3)=x^3-〖5x〗^2+x+6For a cubic curve, the y-intercept must always...

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