Advertisements
Advertisements
प्रश्न
f(x) = \[- (x - 1 )^3 (x + 1 )^2\] .
Advertisements
उत्तर
\[\text { Given:} f\left( x \right) = - \left( x - 1 \right)^3 \left( x + 1 \right)^2 \]
\[ \Rightarrow f'\left( x \right) = - \left[ 3 \left( x - 1 \right)^2 \left( x + 1 \right)^2 + 2\left( x + 1 \right) \left( x - 1 \right)^3 \right]\]
\[\text { For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow - 3 \left( x - 1 \right)^2 \left( x + 1 \right)^2 - 2\left( x + 1 \right) \left( x - 1 \right)^3 = 0\]
\[ \Rightarrow \left( x - 1 \right)^2 \left( x + 1 \right)\left[ - 3\left( x + 1 \right) - 2\left( x - 1 \right) \right] = 0\]
\[ \Rightarrow \left( x - 1 \right)^2 \left( x + 1 \right)\left[ - 3x - 3 - 2x + 2 \right] = 0\]
\[ \Rightarrow \left( x - 1 \right)^2 \left( x + 1 \right)\left[ - 5x - 1 \right] = 0\]
\[ \Rightarrow x = 1, - 1 \text { and }\frac{- 1}{5}\]
\[\text { Thus, x = 1, x = - 1 and } x = \frac{- 1}{5} \text { are the possible points of local maxima or local minima }. \]
\[\text { Now,} \]
\[f''\left( x \right) = - \left[ 3\left\{ 2\left( x - 1 \right) \left( x + 1 \right)^2 + 2\left( x + 1 \right) \left( x - 1 \right)^2 \right\} + 2\left\{ \left( x - 1 \right)^3 + 3 \left( x - 1 \right)^2 \left( x + 1 \right) \right\} \right]\]
\[ = - 6\left( x - 1 \right) \left( x + 1 \right)^2 + 6\left( x + 1 \right) \left( x - 1 \right)^2 - 2 \left( x - 1 \right)^3 - 6 \left( x - 1 \right)^2 \left( x + 1 \right)\]
\[\text { At x} = 1: \]
\[ f''\left( 1 \right) = - 6\left( 1 - 1 \right) \left( 1 + 1 \right)^2 + 6\left( 1 + 1 \right) \left( 1 - 1 \right)^2 - 2 \left( 1 - 1 \right)^3 - 6 \left( 1 - 1 \right)^2 \left( 1 + 1 \right) = 0\]
\[\text { So, it is a point of inflexion } . \]
\[\text { At } x = - 1: \]
\[ f''\left( - 1 \right) = - 6\left( - 1 - 1 \right) \left( - 1 + 1 \right)^2 + 6\left( - 1 + 1 \right) \left( - 1 - 1 \right)^2 - 2 \left( - 1 - 1 \right)^3 - 6 \left( - 1 - 1 \right)^2 \left( - 1 + 1 \right) = 16 > 0\]
\[\text{ So, x = - 1 is the point of local minimum }. \]
\[\text { The local minimum value is given by } \]
\[f\left( - 1 \right) = - \left( 1 - 1 \right)^3 \left( - 1 + 1 \right)^2 = 0\]
\[\text { At } x = - \frac{1}{5}: \]
\[ f''\left( - \frac{1}{5} \right) = - 6\left( - \frac{1}{5} - 1 \right) \left( - \frac{1}{5} + 1 \right)^2 + 6\left( - \frac{1}{5} + 1 \right) \left( - \frac{1}{5} - 1 \right)^2 + 2 \left( - \frac{1}{5} - 1 \right)^3 - 6 \left( - \frac{1}{5} - 1 \right)^2 \left( - \frac{1}{5} + 1 \right)\]
\[ = \frac{576}{125} + \frac{384}{125} - \frac{432}{125} - \frac{864}{125} = \frac{- 336}{125} < 0\]
\[\text { So,} x = - \frac{1}{5} \text { is the point of local maximum }. \]
\[\text { The local maximum value is given by }\]
\[f\left( - \frac{1}{5} \right) = - \left( - \frac{1}{5} - 1 \right)^3 \left( - \frac{1}{5} + 1 \right)^2 = - \left( \frac{- 216}{125} \right)\left( \frac{16}{25} \right) = \frac{3465}{3125}\]
APPEARS IN
संबंधित प्रश्न
f(x) = - (x-1)2+2 on R ?
f(x)=| x+2 | on R .
f (x) = \[-\] | x + 1 | + 3 on R .
f(x) = (x \[-\] 1) (x+2)2.
f(x) = \[\frac{1}{x^2 + 2}\] .
f(x) = cos x, 0 < x < \[\pi\] .
f(x) = (x - 1) (x + 2)2.
f(x) = \[x^3 - 2a x^2 + a^2 x, a > 0, x \in R\] .
`f(x)=xsqrt(1-x), x<=1` .
The function y = a log x+bx2 + x has extreme values at x=1 and x=2. Find a and b ?
Show that \[\frac{\log x}{x}\] has a maximum value at x = e ?
`f(x) = 3x^4 - 8x^3 + 12x^2- 48x + 25 " in "[0,3]` .
Find the absolute maximum and minimum values of a function f given by `f(x) = 12 x^(4/3) - 6 x^(1/3) , x in [ - 1, 1]` .
Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm3, which has the minimum surface area?
Two sides of a triangle have lengths 'a' and 'b' and the angle between them is \[\theta\]. What value of \[\theta\] will maximize the area of the triangle? Find the maximum area of the triangle also.
A tank with rectangular base and rectangular sides, open at the top, is to the constructed so that its depth is 2 m and volume is 8 m3. If building of tank cost 70 per square metre for the base and Rs 45 per square metre for sides, what is the cost of least expensive tank?
A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimension of the rectangular of the window to admit maximum light through the whole opening.
Show that the height of the cylinder of maximum volume that can be inscribed a sphere of radius R is \[\frac{2R}{\sqrt{3}} .\]
A rectangle is inscribed in a semi-circle of radius r with one of its sides on diameter of semi-circle. Find the dimension of the rectangle so that its area is maximum. Find also the area ?
Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm ?
Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius \[5\sqrt{3 cm} \text { is }500 \pi {cm}^3 .\]
Find the point on the curve y2 = 4x which is nearest to the point (2,\[-\] 8).
Find the point on the curve x2 = 8y which is nearest to the point (2, 4) ?
Find the point on the curvey y2 = 2x which is at a minimum distance from the point (1, 4).
An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. Show that the expenses of lining with lead with be least, if depth is made half of width.
Write the minimum value of f(x) = \[x + \frac{1}{x}, x > 0 .\]
Write the point where f(x) = x log, x attains minimum value.
For the function f(x) = \[x + \frac{1}{x}\]
Let f(x) = x3+3x2 \[-\] 9x+2. Then, f(x) has _________________ .
The minimum value of f(x) = \[x4 - x2 - 2x + 6\] is _____________ .
The number which exceeds its square by the greatest possible quantity is _________________ .
If x lies in the interval [0,1], then the least value of x2 + x + 1 is _______________ .
If x+y=8, then the maximum value of xy is ____________ .
If(x) = x+\[\frac{1}{x}\],x > 0, then its greatest value is _______________ .
Let f(x) = 2x3\[-\] 3x2\[-\] 12x + 5 on [ 2, 4]. The relative maximum occurs at x = ______________ .
