Question

Find the x values at which following functions attain maxima and minima 1) f(x)=2x^3–3x^2–12x+5. 2) f(x)=x^3–3x.

Answer 1

1)

Given:

A function

f(x)= 2x³-3x²-12x+5

To Find:

Values of x at which function attains the maxima and minima

Solution:

We know that,

• For the given function f(x)

✏ Solutions of f'(x)=0 are points of maxima and minima of f(x)

✏ If f''(x₁) is positive or equal to zero, then x₁ is the minima  

✏ If f''(x₁) is negative, then x₁ is the maxima

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On differentiating f(x) w.r.t x, we get

$\longrightarrow\rm{f'(x)=2(3x^{2})-3(2x)-12(1)+0}$

$\longrightarrow\rm{f'(x)=6x^{2}-6x-12}$

Now, Let f'(x)=0

$\longrightarrow\rm{f'(x)=0}$

$\longrightarrow\rm{6x^{2}-6x-12=0}$

$\longrightarrow\rm{6(x^{2}-x-2)=0}$

$\longrightarrow\rm{x^{2}-x-2=0}$

$\longrightarrow\rm{x^{2}-2x+x-2=0}$

$\longrightarrow\rm{x(x-2)+1(x-2)=0}$

$\longrightarrow\rm{(x+1)(x-2)=0}$

So, the solutions of above equation are

$\longrightarrow\rm{x=-1,2}$

Therefore, -1 and 2 are the points to be checked for maxima and minima

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On differentiating f'(x) w.r.t. x, we get

$\longrightarrow\rm{f''(x)=6(2x)-6(1)-0}$

$\longrightarrow\rm{f''(x)=12x-6}$

On putting x=-1 in in f''(x), we get

$\longrightarrow\rm{f''(-1)=12(-1)-6}$

$\longrightarrow\rm{f''(-1)=-12-6}$

$\longrightarrow\rm{f''(-1)=-18}$

$\longrightarrow\rm{f''(-1)<0}$

So, x= -1 is the point of maxima

Now,

On putting x=2 in in f''(x), we get

$\longrightarrow\rm{f''(2)=12(2)-6}$

$\longrightarrow\rm{f''(2)=24-6}$

$\longrightarrow\rm{f''(2)=18}$

$\longrightarrow\rm{f''(2)>0}$

So, x= 2 is the point of minima

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2)

Given:

A function

f(x)= x³-3x

To Find:

Values of x at which function attains the maxima and minima

Solution:

We know that,

• For the given function f(x)

✏ Solutions of f'(x)=0 are points of maxima and minima of f(x)

✏ If f''(x₁) is positive or equal to zero, then x₁ is the minima  

✏ If f''(x₁) is negative, then x₁ is the maxima

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On differentiating f(x) w.r.t x, we get

$\longrightarrow\rm{f'(x)=3x^{2}-3(1)}$

$\longrightarrow\rm{f'(x)=3x^{2}-3}$

Now, Let f'(x)=0

$\longrightarrow\rm{f'(x)=0}$

$\longrightarrow\rm{3x^{2}-3=0}$

$\longrightarrow\rm{3(x^{2}-1)=0}$

$\longrightarrow\rm{x^{2}-1=0}$

$\longrightarrow\rm{(x+1)(x-1)=0}$

So, the solutions of above equation are

$\longrightarrow\rm{x=-1,1}$

Therefore, -1 and 1 are the points to be checked for maxima and minima

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On differentiating f'(x) w.r.t. x, we get

$\longrightarrow\rm{f''(x)=3(2x)-0}$

$\longrightarrow\rm{f''(x)=6x}$

On putting x=-1 in in f''(x), we get

$\longrightarrow\rm{f''(-1)=6(-1)}$

$\longrightarrow\rm{f''(-1)=-6}$

$\longrightarrow\rm{f''(-1)<0}$

So, x= -1 is the point of maxima

On putting x=1 in in f''(x), we get

$\longrightarrow\rm{f''(1)=6(1)}$

$\longrightarrow\rm{f''(1)=6}$

$\longrightarrow\rm{f''(1)>0}$

So, x= 1 is the point of minima

Answer 2

Answer

To find maxima and minima :-

Step - 1

Find $\frac{dy}{dx} = {f}^{'}$ and put it equal to zero.

$\frac{dy}{dx}$ = 0 , by solving it we will get some values of x ( let $x_1 and x_2$ )

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Step - 2

Find $\frac{ {d}^{2} y}{ {dx}^{2} } = {f}^{"}$ and put the values of x found in previous step.

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Check

If$\frac{ {d}^{2} y}{ {dx}^{2} } > 0$ for any x = $x_2$ , then $x_2$Is maxima.

If$\frac{ {d}^{2} y}{ {dx}^{2} } < 0$ for any x = $x_1$, then $x_1$ Is minima.

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1.

Equation - $2 {x}^{3} - 3 {x}^{2} - 12x + 5$

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Differentiation of equation = $2 {x}^{3} - 3 {x}^{2} - 12x + 5$

$= 6 {x}^{2} - 6x - 12$

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Putting the equation = 0

$6 {x}^{2} - 6x - 12 = 0$

$\longrightarrow$$= {x}^{2} - x - 2 = 0$

$\longrightarrow$$= {x}^{2} - 2x + x - 2 = 0$

$\longrightarrow$$= (x + 1)(x - 2) = 0$

Roots of the equation are -1 , 2

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${f}^{"}$= Differentiation of $6 {x}^{2} - 6x - 12$

${f}^{"} = 12x - 6$

Substituting the roots of the equation in ${f}^{"}$

12x - 6 = -12-6 = -18

12x - 6 = 12 × 2 - 6 = 24 - 6 = 18

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$\longrightarrow$At x = -1 , the value of function is negative so , x = -1 is the maxima.

$\longrightarrow$At x = 2 , the value of function is positive so , x = 2 is the minima.

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2. Equation - ${x}^{3} - 3 x$

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Differentiation of equation = ${x}^{3} - 3 {x}$

$\longrightarrow$$= 3 {x}^{2} - 3$

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Putting the equation = 0

$\longrightarrow$$3 {x}^{2} - 3 = 0$

$\longrightarrow$$= {x}^{2} - 1 = 0$

$\longrightarrow$$= (x + 1)(x - 1) = 0$

Roots of the equation are -1 , +1

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${f}^{"}$= Differentiation of ${x}^{2} - 3$

${f}^{"} = 6x$

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Substituting the roots of the equation in ${f}^{"}$

$\longrightarrow$2x = -6 at x = -1

$\longrightarrow$2x = 6 at x = 1

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$\longrightarrow$At x = -1 , the value of function is negative so , x = -1 is the maxima.

$\longrightarrow$and At x = 1 , the value of function is positive so , x = 1 is the minima.

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Thanks