(a) \(f(x)=x^4-2x^3+x^2-5x+7\)
(b) \(f(x)=\frac{\textstyle 1}{\textstyle x^2}+\sqrt{x^3}\)
(c) \(f(x)=2e^x+\ln x+\cos x\)
Solutions: (a) \(f'(x)=4x^3-6x^2+2x-5\); (b) \(f'(x)=-\frac{2}{x^3}+\frac{3}{2}\sqrt{x}\); (c) \(f'(x)=2e^x+\frac{1}{x}-\sin x\)(a) \(f(x)=x^3 \ln x\)
(b) \(f(x)=e^x \sin x\)
(c) \(f(x)=\frac{\textstyle 2x+1}{\textstyle x-2}\)
(d) \(f(x)=\frac{\textstyle \sin x+2}{\textstyle \cos x}\)
Solutions: (a) \(f'(x)=3x^2 \ln x+x^2\); (b) \(f'(x)=e^x \sin x +e^x \cos x\); (c) \(f'(x)=\frac{-5}{x^2-4x+4}\); (d) \(f'(x)=\frac{1+2\sin x}{\cos^2 x}\)(a) \(y=(4x+1)^3\)
(b) \(y=\sqrt{x^2+9}\)
(c) \(y=\sin\frac{\textstyle x+\pi}{\textstyle 5}\)
Solutions: (a) \(y'=3\,(4x+1)^2\cdot4=12\,(4x+1)^2\); (b) \(y'=\frac{x}{\sqrt{x^2+9}}\); (c) \(y'=\frac{1}{5}\cos\frac{x+\pi}{5}\)(a) Differentiate the function \(f\).
(b) Write down the equation of the tangent at \(x=4\).
(c) Show that this tangent is parallel to the straight line \(y=3x+1\).
Solutions: (a) \(f'(x)=\frac{2x+1}{3}\); (b) tangent: \(y=3x-5\); (c) they have the same gradient: \(m_1=m_2=3\)(a) Differentiate the function \(f\).
(b) Write down the equation of the normal at \(x=1\).
(c) This normal and both coordinate axes form a triangle:
(i) Write down the coordinates of the vertices of this triangle.
(ii) Calculate the area of this triangle.
Solutions: (a) \(f'(x)=\frac{1}{\sqrt{2x+7}}\); (b) normal: \(y=-3x+6\); (c) (i) vertices: \(A(0,0),~ B(2,0),~ C(0,6)\); (ii) area: \(A=6\)(a) Find the zeros of this function.
(b) Find the stationary points of this function.
(c) Draw the graph.
Solutions: (a) \(x_1=-2,~ x_{2,3}=0,~ x_4=2\); (b) \(P_1(-\sqrt{2},-4),~ P_2(0,0),~ P_3(\sqrt{2},-4)\)(a) Find the stationary points of this function.
(b) Draw the graph using your GDC and verify the obtained result.
Solutions: (a) \(P_1(-1,1)\)(a) Find the stationary points of this function.
(b) Draw the graph using your GDC and verify the obtained result.
Solutions: (a) \(P_1(-2,-2),~ P_2(2,2)\)(a) \(f(x)=x^3+2x^2-5x+4\)
(b) \(f(x)=\frac{\textstyle x^2+3x+1}{\textstyle 5}\)
(c) \(f(x)=\sqrt{x^5}\)
Solutions: (a) \(f'(x)=3x^2+4x-5,~ f''(x)=6x+4\); (b) \(f'(x)=\frac{2x+3}{5},~ f''(x)=\frac{2}{5}\); (c) \(f'(x)=\frac{5}{2}\sqrt{x^3},~ f''(x)=\frac{15}{4}\sqrt{x}\)(a) \(y=e^x+\sin x\)
(b) \(y=x-\ln x\)
(c) \(y=x^2 \ln x\)
Solutions: (a) \(y'=e^x+\cos x,~ y''=e^x-\sin x\); (b) \(y'=1-\frac{1}{x},~ y''=\frac{1}{x^2}\); (c) \(y'=2x\ln x+x,~ y''=2\ln x+3\)(a) \(y=e^x\sin x\)
(b) \(y=\frac{\textstyle 1}{\textstyle x+3}\)
Solutions: (a) \(\frac{\mathrm{d}y}{\mathrm{d}x}=e^x\sin x+e^x\cos x,~ \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=2e^x\cos x\); (b) \(\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{(x+3)^2},~ \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{2}{(x+3)^3}\)(a) \(y=x^2+ax+b\)
(b) \(y=\frac{\textstyle x}{\textstyle a}+\frac{\textstyle b}{\textstyle x}\)
Solutions: (a) \(y'=2x+a,~ y''=2\); (b) \(y'=\frac{1}{a}-\frac{b}{x^2},~ y''=\frac{2b}{x^3}\)(a) Using the first derivative find the stationary points.
(b) Using the second derivative determine the type of each of these points.
Solutions: Maximum: \(P_1(0,1)\), minimum: \(P_2(2,-3)\)(a) Find the stationary points.
(b) Determine the nature of each stationary point.
Solutions: Minimum: \(P_1(-\sqrt{2},-4)\), maximum: \(P_2(0,0)\), minimum: \(P_3(\sqrt{2},-4)\)(a) Find the stationary points.
(b) Determine the nature of each stationary point.
(c) Using your GDC draw the graph to verify the obtained results.
Solutions: Minimum: \(P_1(-1,1)\), inflexion point: \(P_2(0,3)\)(a) Find the zeros, vertical asymptotes and horizontal asymptote.
(b) Find the stationary points and determine their types.
(c) Hence draw the graph.
Solutions: (a) Zeros: \(x_1=\frac{1}{2}\), vertical asymptote: \(x=0\), horizontal asymptote: \(y=0\); (b) maximum: \(P_1(1,1)\)(a) Find the zeros, vertical asymptotes and horizontal asymptote.
(b) Find and classify the stationary points.
(c) Hence draw the graph.
Solutions: (a) Zeros: \(x_1=0,~ x_2=3\), no vertical asymptotes, horizontal asymptote: \(y=1\); (b) maximum: \(P_1(-3,\frac{3}{2})\), minimum: \(P_2(1,-\frac{1}{2})\)(a) Find the zeros and asymptotes, if any.
(b) Find and classify the stationary points.
(c) Hence draw the graph.
Solutions: (a) No zeros, no vertical asymptotes, horizontal asymptote: \(y=0\); (b) maximum: \(P_1(2,3)\)(a) Find and classify the stationary points.
(b) Using your GDC draw the graph to verify the obtained results.
Solutions: (a) Maximum: \(P_1(1,1)\)(a) \(y=x^3-3x^2+3x\)
(b) \(y=x^3-3x^2+2x+1\)
(c) \(y=x^3-3x^2+4x-1\)
Solutions: In all three cases \(y''=0\) at \(P_1(1,1)\). These points are the inflexion points.(a) Find zeros.
(b) Find and classify the stationary points.
(c) Find and classify the inflexion points.
(d) Hence draw the graph.
Solutions: (a) Zeros: \(x_1=0,~ x_{2,3}=3\); (b) maximum: \(P_1(1,4)\), minimum: \(P_2(3,0)\); (c) non-stationary inflexion point: \(P_3(2,2)\)(a) Find zeros.
(b) Find and classify the stationary points.
(c) Find and classify the inflexion points.
(d) Hence draw the graph.
Solutions: (a) Zeros: \(x_{1,2,3}=0,~ x_4=-4\); (b,c) minimum: \(P_1(-3,-3)\), stationary inflexion point: \(P_2(0,0)\), non-stationary inflexion point: \(P_3(-2,-\frac{16}{9})\)(a) Find the inflexion points.
(b) Write the equation of the tangent at the inflexion point with the negative abscissa.
Solutions: (a) Inflexion points (both non-stationary): \(P_1(-1,3),~ P_2(1,3)\); (b) tangent: \(y=\frac{3}{2}x+\frac{9}{2}\)