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