Question

Question :
Determine all the points of local maxima and local minima of the following function : f(x)=(-3/4)x⁴-8x³-(45/2)x² +105

Thank you :)​

Answer 1

$\huge{ \underline{ \underline{ \sf {Required \: answer -: }}}}$

$</p><p>\bf{f(x)=-\dfrac{3x^{4}}{4}-8x^{3}-\dfrac{45x^{2}}{2}+105}$

To Find:

• All the points of local maxima and local minima of the given function.

Solution:

We know that,

For the given function f(x)

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

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

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

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

$\rm{f'(x)=-\dfrac{12x^{3}}{4}-24x^{2}-\dfrac{90x}{2}+0}$

$\longrightarrow\rm{f'(x)=-\dfrac{12x^{3}}{4}-24x^{2}-\dfrac{90x}{2}}$

$\longrightarrow\rm{f'(x)=0}⟶f$

$\longrightarrow\rm{-\dfrac{12x^{3}}{4}-24x^{2}-\dfrac{90x}{2}=0}$

$\longrightarrow\rm{-2\bigg(\dfrac{6x^{3}}{4}+12x^{2}+\dfrac{45x}{2}\bigg)=0}$

$\longrightarrow\rm{\dfrac{6x^{3}}{4}+12x^{2}+\dfrac{45x}{2}=0}$

$\longrightarrow\rm{\dfrac{6x^{3}+48x^{2}+90x}{4}=0}$

$\longrightarrow\rm{6x^{3}+48x^{2}+90x=0}$

$\longrightarrow\rm{6(x^{3}+8x^{2}+15x)=0}$

$\longrightarrow\rm{x^{3}+8x^{2}+15x=0}$

$\longrightarrow\rm{x(x^{2}+8x+15)=0}$

$\longrightarrow\rm{x(x^{2}+5x+3x+15)=0}$

$\longrightarrow\rm{x(x(x+5)+3(x+5))=0}$

$\longrightarrow\rm{x((x+3)(x+5))=0}$

$\longrightarrow\rm{x(x+3)(x+5)=0}$

So, the solutions of above equation are

$\rm{x=0,-3,-5}⟶x=0,−3,−5$

Therefore, 0,-3 and -5 are the points to be checked for local maxima and minima

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

$\longrightarrow\rm{f''(x)=-\dfrac{36x^{2}}{4}-48x-\dfrac{90}{2}}$

$\longrightarrow\rm{f''(x)=-9x^{2}-48x-45}$

On putting x=0 in in f''(x), we get

$\longrightarrow\rm{f''(0)=-9(0)^{2}-48(0)-45}$

$\longrightarrow\rm{f''(0)=-45}$

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

So, x=0 is the point of local maxima

Now,

On putting x= -3 in in f''(x), we get

$\longrightarrow\rm{f''(-3)=-9(-3)^{2}-48(-3)-45}$

$\longrightarrow\rm{f''(-3)=-81+144-45}$

$\longrightarrow\rm{f''(-3)=18}$

$\longrightarrow\rm{f''(-5)=-9(-5)^{2}-48(-5)-45}$

So, x= -3 is the point of local minima

Now,

On putting x= -5 in in f''(x), we get,

$\rm{f''(-5)=-9(-5)^{2}-48(-5)-45}$

$\longrightarrow\rm{f''(-5)=-225+240-45}$

$\longrightarrow\rm{f''(-5)=-30}$

$\longrightarrow\rm{f''(-5) < 0}$

So, x= -5 is the point of local maxima.