The value of the function f(x) = (x - 1)(x - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x) = (x - 1)(x - 2)^2$. Expanding the function: $f(x) = (x - 1)(x^2 - 4x + 4) = x^3 - 5x^2 + 8x - 4$. To find the critical points,we calc

Vedclass · Accepted Answer

$\frac{4}{27}$

Vedclass · Answer

(D) Given $f(x) = (x - 1)(x - 2)^2$. Expanding the function: $f(x) = (x - 1)(x^2 - 4x + 4) = x^3 - 5x^2 + 8x - 4$. To find the critical points,we calculate the first derivative: $f'(x) = 3x^2 - 10x + 8$. Setting $f'(x) = 0$: $3x^2 - 10x + 8 = 0$,which factors as $(3x - 4)(x - 2) = 0$. Thus,the critical points are $x = \frac{4}{3}$ and $x = 2$. Now,we find the second derivative: $f''(x) = 6x - 10$. Evaluating at $x = \frac{4}{3}$: $f''(\frac{4}{3}) = 6(\frac{4}{3}) - 10 = 8 - 10 = -2 Evaluating at $x = 2$: $f''(2) = 6(2) - 10 = 2 > 0$. The function has a local minimum at $x = 2$. The maximum value is $f(\frac{4}{3}) = (\frac{4}{3} - 1)(\frac{4}{3} - 2)^2 = (\frac{1}{3})(-\frac{2}{3})^2 = (\frac{1}{3})(\frac{4}{9}) = \frac{4}{27}$.

The value of the function $f(x) = (x - 1)(x - 2)^2$ at its maxima is

Similar Questions

The perimeter of a sector is a constant. If its area is to be maximum,then the sectorial angle is

The minimum value of $f(x) = e^{(x^4 - x^3 + x^2)}$ is

If $\overline{PQ}$ and $\overline{PR}$ are two sides of $\Delta PQR$,for what value of the angle between $\overline{PQ}$ and $\overline{PR}$ is the area of the triangle maximum?

Find the minimum value of $2x + 3y$ when $xy = 6$.

The number of real roots of the equation $e^{4x} + 2e^{3x} - e^{x} - 6 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module