7. Find both the maximum value and the minimum value of
3x^4 - 8x^3 + 12x^2 - 48x + 25 on the interval [0, 3].

Answers (1)

Given function is
f(x)=3x^4 - 8x^3 + 12x^2 - 48x + 25^{}
f^{'}(x)=12x^3 - 24x^2 +24x - 48 \\ f^{'}(x)=0\\ 12(x^3-2x^2+2x-4) = 0\\ x^3-2x^2+2x-4=0\\
Now, by hit and trial let first assume x = 2
(2)^3-2(2)^2+2(2)-4\\ 8-8+4-4=0
Hence, x = 2 is one value
Now,
\frac{x^3-2x^2+2x-4}{x-2} = \frac{(x^2+2)(x-2)}{(x-2)} = (x^2+2)
x^2 = - 2   which is  not possible 
Hence, x = 2 is the only critical value  of function f(x)=3x^4 - 8x^3 + 12x^2 - 48x + 25^{}
Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 0 and x = 3
f(2)=3(2)^4 - 8(2)^3 + 12(2)^2 - 48(2) + 25^\\
           =3\times16 - 8\times 8 + 12\times 4 - 96 + 25 = 48-64+48-96+25 = -39

f(3)=3(3)^4 - 8(3)^3 + 12(3)^2 - 48(3) + 25\\ =3\times81-8\times27+12\times9-144+25 \\ =243-216+108-144+25 = 16

f(0)=3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 25
Hence, maximum value of function  f(x)=3x^4 - 8x^3 + 12x^2 - 48x + 25^{}  occurs at x = 0 and vale is 25
and minimum value of function f(x)=3x^4 - 8x^3 + 12x^2 - 48x + 25^{} occurs at x = 2 and value is -39

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions