Question

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (i). f(x) = x² (ii). g(x) = x³ − 3x (iii). h(x) = sinx + cos, 0 <x<$\frac{\pi}{2}$ (iv). f(x) = sinx − cos x, 0 < x < 2π (v). f(x) = x³ − 6x² + 9x + 15(vi) g(x)=$\frac{x}{2}$+$\frac{2}{x}$>0 (vii).g(x)=$\frac{1}{x^2}$+2(viii). f(x)=x√1-x,x>0

Answer 1

(i) f(x) = x2

∴f'(x)=0=x=0

Thus, x = 0 is the only critical point that could possibly be the point of local maxima or local minima of f. We have f''(0)=2, which is positive.

Therefore, by the second derivative test, x = 0 is a point of local minima and the local minimum value of f at x = 0 is f(0) = 0. (ii) g(x) = x3 − 3x

∴ g'(x)=3$\times$2-3

Now

g'(x)=0=3$\times$2=3=x=±1

g'(x)=6x

g'(1)=6>0

g'(-1)=-6<0

By second derivative test, x = 1 is a point of local minima and local minimum value of g at x = 1 is g(1) = 13 − 3 = 1 − 3 = −2. However, x = −1 is a point of local maxima and local maximum value of g at x = −1 is g(1) = (−1)3 − 3 (− 1) = − 1 + 3 = 2.

(iii) h(x) = sinx + cosx, 0 < x <$\frac{\pi}{2}$

h'(x)=cos x-sinx

h'(x)=0=sinx=cosx=tanx₁=x=$\frac{\pi}{4}$∈(0,$\frac{\pi}{2}$)

h$\times$(X)=-sinx-cosx=-(sinx+cos x)

h$\times$(π/4)=-($\frac{1}{\sqrt 2}$+$\frac{1}{\sqrt 2}$)=$\frac{2}{\sqrt 2}$=√2<0.

Therefore, by second derivative test, x=$\frac{\pi}{4}$ is a point of local maxima and the local

maximum value of h at is x=$\frac{\pi}{4}$ is h($\frac{\pi}{4}$)=sin +cos $\frac{\pi}{4}$=$\frac{1}{\sqrt 2}$+$\frac{1}{\sqrt 2}$=√2. 

(iv) f(x) = sin x − cos x, 0 < x < 2π

f'(x)=cosx+sinx

f'(x)=0=cosx=-sin x=tanx=-1=x=$\frac{3\pi}{4}$,$\frac{7\pi}{4}$∈(0,2π)

f*(x)-sinx+cosx

f''(3π/4)=-sin 3π/4+cos 3π/4=-$\frac{1}{\sqrt 2}$-$\frac{1}{\sqrt 2}$=-√2>0

f''(7π/4)=-sin 7π/4+cos 7π/4=-$\frac{1}{\sqrt 2}$-$\frac{1}{\sqrt 2}$=-√2>0

Therefore, by second derivative test x=3π/4, is a point of local maxima and the local maximum value of f at x=$\frac{3\pi}{4}$, is -√2

the local minimum value of f at is.

(v) f(x) = x³−6x²+9x+15

f'(x)=3x²-12x+9

x=1,3

Now, f''

(x)=6x-12=6(x-2)

f$\times$(1)=6(1-2)=-6<0

f$\times$(3)=6(1-2)=-6>0

Therefore, by the second derivative test, x = 1 is a point of local maxima and the local maximum value of f at x = 1 is f(1) = 1 − 6 + 9 + 15 = 19. However, x = 3 is a point of local minima and the local minimum value of f at x = 3 is f(3) = 27 − 54 + 27 + 15 = 15

g$\times$(2)=$\frac{4}{23}$=$\frac{1}{2}$>0

Therefore, by the second derivative test, x = 2 is a point of local minima and the local minimum value of g at x = 2 is g(2) =$\frac{2}{2}$+$\frac{2}{2}$=1+1=2.

∴f($\frac{2}{3}$)=$\frac{2}{3}$√I-$\frac{2}{3}$=$\frac{2}{3}$√I/3=2/3