y=x6-5x5+5x4-10 find minimum and maximum value
Answers
Answered by
1
Step-by-step explanation:
Given : Equation x^5-5x^4+5x^3-10x
5
−5x
4
+5x
3
−10
To find : The minimum and maximum values of the equation ?
Solution :
To find the maximum minimum points we find the first derivative and get the critical points then substitute that points in the second derivative.
Let y=x^5-5x^4+5x^3-10y=x
5
−5x
4
+5x
3
−10
Derivate w.r.t x,
y'=5x^4-20x^3+15x^2y
′
=5x
4
−20x
3
+15x
2
For critical points put y'=0,
5x^4-20x^3+15x^2=05x
4
−20x
3
+15x
2
=0
5x^2(x^2-4x+3)=05x
2
(x
2
−4x+3)=0
5x^2(x^2-3x-x+3)=05x
2
(x
2
−3x−x+3)=0
5x^2(x(x-3)-1(x-3))=05x
2
(x(x−3)−1(x−3))=0
5x^2(x-1)(x-3)=05x
2
(x−1)(x−3)=0
5x^2=0,x-1=0,x-3=05x
2
=0,x−1=0,x−3=0
x=0,1,3x=0,1,3
The second derivative is
y''=20x^3-60x^2+30xy
′′
=20x
3
−60x
2
+30x
Substitute the critical values,
At x=0,
y''=20(0)^3-60(0)^2+30(0)y
′′
=20(0)
3
−60(0)
2
+30(0)
y''=0y
′′
=0
There is no maximum and no minimum value.
At x=1,
y''=20(1)^3-60(1)^2+30(1)y
′′
=20(1)
3
−60(1)
2
+30(1)
y''=20-60+30y
′′
=20−60+30
y''=-10 < 0y
′′
=−10<0
The value is maximum.
At x=0,
y''=20(3)^3-60(3)^2+30(3)y
′′
=20(3)
3
−60(3)
2
+30(3)
y''=20(27)-60(9)+90y
′′
=20(27)−60(9)+90
y''=540-540+90y
′′
=540−540+90
y''=90 > 0y
′′
=90>0
The value is minimum.
Hope it helps :)
Mark me as a brainliest!
Thank my answer ❤️
Answered by
0
Step-by-step explanation:
Answer
To obtain the absolute maxima or minima for the function f(x)
Find f
′
(x) and put f
′
(x)=0
Given
x
5
−5x
4
+5x
3
−10
f(x)=x
5
−5x
4
+5x
3
−10
f(x)=x
5
−5x
4
+5x
3
−10
dx
df(x)
=f
′
(x)=5x
4
−20x
3
+15x
2
5x
4
−20x
3
+15x
2
=0
5x
2
(x−3)(x−1)=0
x=0, x=1, x=3
To find the maximum find f
′′
(x) and substitute the value of x
dx
df
′
(x)
=f
′′
(x)=20x
3
−60x
2
+30x
When, x=3
f
′′
(x)=20x
3
−60x
2
+30x=20(3)
3
−60(3)
2
+30(3)=90 >0
When, x=0
f
′′
(x)=20x
3
−60x
2
+30x=20(0)
3
−60(0)
2
+30(0)=0
When, x=1
f
′′
(x)=20x
3
−60x
2
+30x=20(1)
3
−60(1)
2
+30(1)=−10 <0
∴ The given function has the maximum value when x=3
Similar questions