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

Given:

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

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

$\rule{190}{1}$

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

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

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

Now, Let f'(x)=0

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

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

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

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

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

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

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

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

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

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

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

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

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

So, the solutions of above equation are

$\longrightarrow\rm{x=0,-3,-5}$

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

$\rule{190}{1}$

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

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

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

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

$\longrightarrow\rm{f''(-3)>0}$

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

Now,

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

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

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

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

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

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

Answer 2

f ′ (x) = –3x3 – 24x2 – 45x

= – 3x (x2 + 8x + 15)

= – 3x (x + 5) (x + 3)

f ′ (x) = 0 ⇒ x = –5, x = –3, x = 0

f ″(x) = –9x2 – 48x – 45

= –3 (3x2 + 16x + 15)

f ″(0) = – 45 < 0. Therefore, x = 0 is point of local maxima

f ″(–3) = 18 > 0. Therefore, x = –3 is point of local minima

f ″(–5) = –30 < 0. Therefore x = –5 is point of local maxima.

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