Find the intervals in which the function $f(x) = 2x^2 - 3x$ is:

(a) strictly increasing
(b) strictly decreasing

Q: Find the intervals in which the function $f(x) = 2x^2 - 3x$ is: (a) strictly increasing (b) strictly decreasing

Explanation: We are given: $f(x) = 2x^2 - 3x$. \[\] Differentiate: \[f'(x) = \dfrac{d}{dx}(2x^2 - 3x) = 4x - 3\] Set $f'(x) = 0$ to find critical point: \[4x - 3 = 0 \Rightarrow x = \dfrac{3}{4}\] Now analyze the sign of $f'(x)$: In the interval $\left(-\infty,\dfrac{3}{4}\right)$, we have $f'(x) = 4x - 3 0$ $\Rightarrow f(x)$ is strictly increasing on $\left(\dfrac{3}{4},\infty\right)$ \[\]Final Answer: (a) $f(x)$ is strictly increasing in $\left(\dfrac{3}{4}, \infty\right)$ (b) $f(x)$ is strictly decreasing in $\left(-\infty, \dfrac{3}{4}\right)$

Question

Find the intervals in which the function $f(x) = 2x^2 - 3x$ is:

(a) strictly increasing
(b) strictly decreasing

Accepted Answer

Explanation: We are given:
$f(x) = 2x^2 - 3x$.

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

Set $f'(x) = 0$ to find critical point:
$$4x - 3 = 0 \Rightarrow x = \dfrac{3}{4}$$

Now analyze the sign of $f'(x)$:

In the interval $\left(-\infty,\dfrac{3}{4}\right)$, we have $f'(x) = 4x - 3 < 0$
$\Rightarrow f(x)$ is strictly decreasing on $\left(-\infty,\dfrac{3}{4}\right)$

In the interval $\left(\dfrac{3}{4},\infty\right)$, we have $f'(x) = 4x - 3 > 0$
$\Rightarrow f(x)$ is strictly increasing on $\left(\dfrac{3}{4},\infty\right)$

Final Answer:

(a) $f(x)$ is strictly increasing in $\left(\dfrac{3}{4}, \infty\right)$

(b) $f(x)$ is strictly decreasing in $\left(-\infty, \dfrac{3}{4}\right)$

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