Answer:
-128
Explanation:
Given, f(x)=2x
3
−21x
2
+36x−20
∴f
′
(x)=6x
2
−42x+36
When f(x) is a maximum or a minimum, f
′
(x)=0
Hence, 6x
2
−42x+36=0
x
2
−7x+6=0
x
2
−6x−x+6=0
x(x−6)−1(x−6)=0
(x−6)(x−1)=0
x=1,6
Again f
′′
(x)=12x−42
=6(2x−7)
Now, when x=1,f
′′
(x)=−30 ....[negative]
And when x=6,f
′′
(x)=30 ....[positive]
Hence, f(x) is maximum for x=1 and minimum for x=6.
The maximum and minimum values of f(x) are
f(1)=2(1)
3
−21(1)
2
+36(1)−20
=2−21+36−20=−3
f(6)=2(6)
2
−21(6)
2
+36(6)−20
=432−756+216−20=−128