Question

Find the intervals in which the function f given by f (x) = 2x² – 3x is
(a) strictly increasing (b) strictly decreasing

Answer 1

given, f(x) = 2x² - 3x

differentiate f(x) with respect to x,

f'(x) = 4x - 3 ------(1)

(a) when f(x) is strictly increasing function :

f'(x) > 0

from equation (1),

4x - 3 > 0 => x > 3/4

e.g., x ∈ (3/4, ∞ )

Therefore, the given function (f) is strictly increasing in interval x ∈ (3/4, ∞ ) .

(b) when f(x) is strictly decreasing function :

f'(x) < 0

from equation (1),

4x -3 < 0 => x < 3/4

e.g., x ∈ (-∞ , 3/4)

Therefore, the given function (f) is strictly decreasing in interval x ∈ (-∞ , 3/4)

mark

Step-by-step explanation:

Answer 2

AnswEr :

Given Expression,

$\sf \: f(x) = 2 {x}^{2} - 3x$

We have to find the intervals for which the given function is

• Strictly Increasing
• Strictly Decreasing

Firstly,let us find the derivative of the above function w.r.t. x,

$\longrightarrow \: \sf \: \dfrac{f(x) }{dx} = \dfrac{d(2 {x}^{2} - 3x) }{dx} \\ \\ \longrightarrow \: \sf \: f'(x) = 4x - 3$

Equating f'(x) = 0 because f'(x) becomes a "constant function" at 0. This would help us find the intervals at which the given function is increasing and decreasing.

$\longrightarrow \sf \: 4x - 3 = 0 \\ \\ \longrightarrow \sf \: 4x = 3 \\ \\ \longrightarrow \underline{ \boxed{\sf \: x = \dfrac{3}{4} }}$

• The point (3/4,0) divides the x-axis into two intervals (-∞,3/4) and (3/4,∞).

Strictly Increasing

• A function is said to be strictly increasing if all values of f'(x) > 0.

Therefore,

$\sf \: f'(x) > 0 \\ \\ \implies \: \sf \: 4x - 3 > 0 \\ \\ \implies \sf \: x > \dfrac{3}{4}$

f(x) is strictly increasing in the interval (3/4,∞)

Strictly Decreasing

• A function is said to be strictly decreasing if f'(x) < 0

Therefore,

$\sf \: f'(x) < 0 \\ \\ \implies \: \sf \: x < \dfrac{3}{4}$

f(x) is strictly decreasing in the interval (-∞,3/4)

To sum it up,

$\begin{array}{|c|c|c|} \cline{1-3}\sf Given \ Function & \sf f'(x) & \sf Intervals \\ \cline{1-3} \sf 2x^2 - 3x & \sf 4x-3 & \sf Increasing (\dfrac{3}{4},\infty) \\ \cline{3-3} & & \sf Decreasing (-\infty, \dfrac{3}{4}) \\ \cline{1-3} \end{array}$