find absolute maximum and absolute minimum of the function f(x)=3x^4-4x^3 on the (-1,2)
Answers
EXPLANATION.
Absolute maximum and absolute minimum of the function,
⇒ f(x) = 3x⁴ - 4x³ on the line x∈(-1,2).
Differentiate the function w.r.t x, we get.
⇒ dy/dx = d(3x⁴ - 4x³)/dx.
⇒ dy/dx = 12x³ - 12x².
Put f'(x) = 0 in equation, we get.
⇒ 12x³ - 12x² = 0.
⇒ 12x²(x - 1) = 0.
⇒ x = 1.
⇒ f'(x) = 1.
Put the value of x of Range.
Put x = -1 in equation, we get.
⇒ f(-1) = 3(-1)⁴ - 4(-1)³.
⇒ f(-1) = 3 + 4.
⇒ f(-1) = 7.
Put x = 2 in equation, we get.
⇒ f(2) = 3(2)⁴ - 4(2)³.
⇒ f(2) = 3(16) - 4(8).
⇒ f(2) = 48 - 32.
⇒ f(2) = 16.
Put x = 1 in equation, we get.
⇒ f(1) = 3(1)⁴ - 4(1)³.
⇒ f(1) = 3 - 4.
⇒ f(1) = -1.
Maximum value of f(x) = 16.
Minimum value of f(x) = -1.
MORE INFORMATION.
How to find local maxima or minima.
Step - 1 : = put f'(x) = 0 solve for values of x, let x = x₁, x₂,....
Step - 2: = Let x = x₁ is now checking for local maxima and minima and calculate: f'(x₁ - h) and f'(x₁ + h).
Step - 3: = If h > 0 and h is very very small then,
(1) = f'(x₁ - h) < 0 and f'(x₁ + h) > 0 then x₁ is point of minimum.
(2) = if f'(x₁ - h) > 0 and f'(x₁ + h) < 0 then x₁ is point of maximum.
(3) = if f'(x₁ - h) & f'(x₁ + h) has same sign then x₁ is neither point of maximum nor point of minimum.
In the same way other values of x = x₂, x₃ are checked separately.
✯ Appropriate Question ✯
- find absolute maximum and absolute minimum of the function f(x)=3x^4-4x^3 on the (-1,2)
✯ Solution ✯
- To decide on the extrema, check for maximums & minimums and compare them to the values at the endpoints.
- First, take the derivative to find the critical points (places of possible maximums and minimums):
- f'(x) = 12x3 - 12x2 = 12x2(x - 1)
- f'(x) = 0 when x = 0 or x = 1.
☯ Check the critical points against the end points:
- f(-1) = 7 (end point)
- f(0) = 0 (critical point)
- f(1) = -1 (critical point)
- f(2) = 16 (end point)