Question

f(x)=5-3x+5x²-x³ x ∈ R,Find the maximum and minimum value of the given function.

Answer 1

f(x) = 5 - 3x + 5x² - x³ , x ∈ R.

first of all differentiate with respect to x

f'(x) = -3 + 10x - 3x²

now, f'(x) = 0.

=> -3 + 10x - 3x² = 0

=> -(3x² - 10x + 3) = 0

=> 3x² - 9x - x + 3 = 0

=> 3x(x - 3) - 1(x - 3) = 0

=> (3x - 1)(x - 3) = 0

=> x = 3 and 1/3

now, again differentiate with respect to x,
f"(x) = 10 -6x

at x = 3 , f"(x) = 10 - 6 × 3 < 0
we know, any function attains maximum at x = a when f"(a) < 0 and f'(a) = 0
here, f(x) attains maximum due to f'(3) = 0 and f"(3) < 0 .

hence, maximum value of f(x) = f(3) = 5 - 3x + 10x² - x³
= 5 - 9 + 90 - 27
= -31 + 90
= 59


again, at x = 1/3 , f'(x) = 10 - 6 × 1/3 > 0
we know, any function attains minimum at x = b when f'(b) = 0 and f"(b) > 0 .
here, at x = 3, f(x) attains minimum because f'(1/3) = 0 and f"(1/3) < 0

hence, minimum value of f(x) = f(1/3) = 5 - 3x + 10x² - x³
= 5 - 3 × 1/3 + 10× 1/3² - 1/3³
= 5 - 1 + 10/9 - 1/27
= 4 + (30 - 1)/27
= 4 + 29/27
= (108 + 29)/27
= 137/27

Answer 2

Dear Student:

F(x)=5-3x+5x²-x³ x ∈ R

For,maximum and minimum

We will find $\frac{df}{dx}$ and then equate to zero.

And then again proceed for second derivative.

See the attachment: