Respuesta :
We have been given a function [tex]f(x)=2x^3+3x^2-336x[/tex]. We are asked to find the interval on which function is increasing and decreasing.
(a). First of all, we will find the critical points of function by equating derivative with 0.
[tex]f'(x)=2(3)x^{2}+3(2)x^1-336[/tex]
[tex]f'(x)=6x^{2}+6x-336[/tex]
[tex]6x^{2}+6x-336=0[/tex]
[tex]x^{2}+x-56=0[/tex]
[tex]x^{2}+8x-7x-56=0[/tex]
[tex](x+8)-7(x+8)=0[/tex]
[tex](x+8)(x-7)=0[/tex]
[tex]x=-8,x=7[/tex]
So [tex]x=-8,7[/tex] are critical points and these will divide our function in 3 intervals [tex](-\infty,-8)U(-8,7)U(7,\infty)[/tex].
Now we will find derivative over each interval as:
[tex]f'(x)=(x+8)(x-7)[/tex]
[tex]f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16[/tex]
Since [tex]f'(9)>0[/tex], therefore, function is increasing on interval [tex](-\infty,-8)[/tex].
[tex]f'(x)=(x+8)(x-7)[/tex]
[tex]f'(1)=(1+8)(1-7)=(9)(-6)=-54[/tex]
Since [tex]f'(1)<0[/tex], therefore, function is decreasing on interval [tex](-8,7)[/tex].
Let us check for the derivative at [tex]x=8[/tex].
[tex]f'(x)=(x+8)(x-7)[/tex]
[tex]f'(8)=(8+8)(8-7)=(16)(1)=16[/tex]
Since [tex]f'(8)>0[/tex], therefore, function is increasing on interval [tex](7,\infty)[/tex].
(b) Since [tex]x=-8,7[/tex] are critical points, so these will be either a maximum or minimum.
Let us find values of f(x) on these two points.
[tex]f(-8)=2(-8)^3+3(-8)^2-336(-8)[/tex]
[tex]f(-8)=1856[/tex]
[tex]f(7)=2(7)^3+3(7)^2-336(7)[/tex]
[tex]f(7)=-1519[/tex]
Therefore, [tex](-8,1856)[/tex] is a local maximum and [tex](7,-1519)[/tex] is a local minimum.
(c) To find inflection points, we need to check where 2nd derivative is equal to 0.
Let us find 2nd derivative.
[tex]f''(x)=6(2)x^{1}+6[/tex]
[tex]f''(x)=12x+6[/tex]
[tex]12x+6=0[/tex]
[tex]12x=-6[/tex]
[tex]\frac{12x}{12}=-\frac{6}{12}[/tex]
[tex]x=-\frac{1}{2}[/tex]
Therefore, [tex]x=-\frac{1}{2}[/tex] is an inflection point of given function.
(a). The function is increasing in the interval: [tex](-\infty,-8)U(7,\infty)[/tex]
The function is decreasing in the interval: [tex](-8,7)[/tex]
(b). The local maximum value is [tex]f(-8)=1856[/tex]
The local minimum value is [tex]f(7)=-1519[/tex]
(c). The inflection point is [tex]x=\frac{-1}{2}[/tex].
We have the given function:
[tex]f(x) = 2x^3 + 3x^2 - 336x[/tex]
Differentiating it with respect to x, we get
[tex]f'(x) = 6x^2 + 6x - 336\\f''(x) = 12x + 6[/tex]
Put first derivative equal to zero
[tex]6x^2 + 6x - 336=0\\x^2 + x - 336=0\\(x+8)(x-7)=0\\x=-8,7[/tex]
(a). Now, the function is increasing in the interval:[tex](-\infty,-8)U(7,\infty)[/tex]
and the function is decreasing in the interval:[tex](-8,7)[/tex]
(b). Now, since  [tex]f''(-8)<0[/tex] the function has local maxima at x=-8 and the local maximum value is [tex]f(-8)=1856[/tex]
Also since [tex]f''(7)>0[/tex] the function has local minima at x=7 and the local minimum value is [tex]f(7)=-1519[/tex]
(c). For the inflection point put second derivative equal to zero .
[tex]f''(x)=0\\0=12x+6\\x=\frac{-1}{2}[/tex]
So, the inflection point is [tex]x=\frac{-1}{2}[/tex].
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