Let [f(x) = 4x^3 - 11x^2 + 8x - 5,,x R].

Question

Let $$f(x) = 4{x^3} - 11{x^2} + 8x - 5,\,x \in R$$. Then f :

Accepted Answer

Given,

$f(x) = 4{x^3} - 11{x^2} + 8x - 5$

$$\therefore$$ $f'(x) = 12{x^2} - 22x + 8$

$= 2(6{x^2} - 11x + 4)$

$= 2(6{x^2} - 8x - 3x + 4)$

$$ = 2\left[ {2x(3x - 4) - 1(3x - 4)} \right]$$

$$ = 2\left[ {(3x - 4)(2x - 1)} \right]$$

$f'(x)$ is positive before $${1 \over 2}$$ means slope of f(x) is positive before $${1 \over 2}$$ and f'(x) is negative after $${1 \over 2}$$ means slope of f(x) is negative after $${1 \over 2}$$.

$$\therefore$$ At $${1 \over 2}$$, f(x) is local maximum

Also, $f'(x)$ is negative before $${4 \over 3}$$ means slope of f(x) is negative before $${4 \over 3}$$ and f'(x) is positive after $${4 \over 3}$$ means slope of f(x) is positive after $${4 \over 3}$$.

$$\therefore$$ At $${4 \over 3},\,f(x)$$ is local minimum

From wavy curve method,

$$f'(x) > 0,\,\forall x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)$$

$$\therefore$$ f(x) increasing in $$x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)$$

$$f'(x) < 0\,\forall x \in \left( {{1 \over 2},{4 \over 3}} \right)$$

$$\therefore$$ f(x) decreasing in $$x \in \left( {{1 \over 2},{4 \over 3}} \right)$$

Solutions · 1