Question

Find the minima and maxima value
3x⁴ - 2x³ - 6x² + 6x + 1​

Answer 1

Answer:

f(x) = y = 4x - x2 + 3

First let us find the first derivative

f'(x) = 4(1) - 2x + 0

f'(x) = 4 - 2x

Let f'(x) = 0

4 - 2x = 0

2 (2 - x) = 0

2 - x = 0

x = 2

Now let us find the second derivative

f''(x) = 0 - 2(1)

f''(x) = -2 < 0 Maximum

To find the maximum value, we have to apply x = 2 in the original function.

f(2) = 4(2) - 22 + 3

f(2) = 8 - 4 + 3

f(2) = 11 - 4

f(2) = 7

Therefore the maximum value is 7 at x = 2. Now let us check this in the graph.

Checking :

y = 4x - x2 + 3

The given function is the equation of parabola.

y = -x² + 4 x + 3

y = -(x² - 4 x - 3)

y = -{ x² - 2 (x) (2) + 2² - 2² - 3 }

y = - { (x - 2)² - 4 - 3 }

y = - { (x - 2)² - 7 }

y = - (x - 2)² + 7

y - 7 = -(x - 2)²

(y - k) = -4a (x - h)²

Here (h, k) is (2, 7) and the parabola is open downward.