Question: Show that $y = 2x^3 + ax^2 + bx + 3$ is always increasing if and only if $a^2 < 6b$.

Step 1. Differentiate:

This is a quadratic in $x$ with leading coefficient $6 > 0$ (upward parabola).

Step 2. "Always increasing" means $\dfrac{dy}{dx} > 0$ for all $x$ (or $\geq 0$ with equality only at isolated points). Since the parabola opens upwards, this requires the quadratic to have no real roots OR a repeated root.

Step 3. Discriminant condition:

Answer: Always increasing $\Leftrightarrow a^2 < 6b$.

Tip: When $f'(x)$ is a quadratic, "always positive" needs two ingredients:

• leading coefficient $> 0$, AND
• discriminant $< 0$ (strict).

For "always non-negative" allow discriminant $\leq 0$.

Question: Show that $f'(x) = (2x+1)e^{x^2 + x}$ is positive for all $x > 0$, without solving any inequality.

Step 1. Inspect each factor for $x > 0$:

• Factor 1: $2x + 1$. For $x > 0$: $2x > 0$, so $2x + 1 > 1 > 0$. Positive. $\checkmark$
• Factor 2: $e^{x^2 + x}$. Exponentials are positive for ALL real exponents. Positive. $\checkmark$

Step 2. Conclude: product of two positive quantities is positive.

Tip: Factor-sign arguments avoid algebraic inequality-solving. Look for:

• $e^{(\text{anything})} > 0$ always
• $a^2 \geq 0$ with equality only at $a = 0$
• $\dfrac{1}{(\cdot)^2} > 0$ where defined
• $\sqrt{(\cdot)} \geq 0$ where defined

If every factor has a definite sign, the product's sign is determined without solving anything.

Question: Find the range of $x$ for which $y = x^3 - \dfrac{5}{2}x^2 - 2x + 3$ is decreasing.

Step 1. Differentiate:

Step 2. "Decreasing" means $\dfrac{dy}{dx} < 0$:

Step 3. Factorise: discriminant $= 25 + 24 = 49 = 7^2$.

So $\dfrac{dy}{dx} = 3(x-2)\left(x + \tfrac{1}{3}\right)$. Since the parabola opens upwards, it is below zero BETWEEN the roots.

Step 4. Sign analysis:

Answer: $y$ is decreasing for $-\dfrac{1}{3} < x < 2$.

Tip: For "where is the function decreasing", solve $f'(x) < 0$. For "where increasing", solve $f'(x) > 0$. Stationary points are the boundary $x$-values.

Question: $y = x^3 + px^2 + qx + 10$ is decreasing only for $3 < x < 7$. Find $p$ and $q$.

Step 1. Differentiate: $\dfrac{dy}{dx} = 3x^2 + 2px + q$.

Step 2. "Decreasing only for $3 < x < 7$" means $\dfrac{dy}{dx} < 0$ has solution set $(3, 7)$. So $\dfrac{dy}{dx} = 0$ at $x = 3$ and $x = 7$:

Step 3. Compare coefficients:

• $x$: $2p = -30 \Rightarrow p = -15$
• constant: $q = 63$

Answer: $p = -15$, $q = 63$.

Verification: $\dfrac{dy}{dx} = 3x^2 - 30x + 63 = 3(x-3)(x-7)$. At $x = 5$ (between roots): $3(2)(-2) = -12 < 0$. $\checkmark$

Tip: "Decreasing only for $a < x < b$" means $f'(x)$ has roots exactly at $a$ and $b$ AND is negative between them. Build $f'(x) = (\text{leading coeff})(x-a)(x-b)$, compare to the differentiated form.

Question: $y = x^2 + 4$. Show $y$ is increasing for $x > 0$. Given $\dfrac{dy}{dt} = 0.4$ unit/s at $x = 2$, find $\dfrac{dx}{dt}$ at that instant.

Step 1. Differentiate: $\dfrac{dy}{dx} = 2x$. For $x > 0$: $\dfrac{dy}{dx} = 2x > 0 \Rightarrow$ increasing. $\checkmark$

Step 2. Chain rule:

Step 3. At $x = 2$: $\dfrac{dy}{dx} = 4$, so:

Answer: $\dfrac{dx}{dt} = 0.1$ unit/s.

Tip: Inc/dec + rate-of-change combos: the inc/dec part is usually a setup for stating $\dfrac{dy}{dx}$ has a definite sign. The chain rule $\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}$ then translates between time-rates.

Question: Show that $f'(x) = \dfrac{3}{2x-1}$ is a decreasing function for $x > \dfrac{1}{2}$.

Step 1. To show $f'(x)$ is decreasing, examine the sign of its derivative — which is $f''(x)$.

Step 2. For $x > \tfrac{1}{2}$: $(2x-1)^2 > 0$, so $\dfrac{6}{(2x-1)^2} > 0$, and:

Step 3. Since $f''(x) < 0$ everywhere on the domain $x > \tfrac{1}{2}$, the gradient function $f'(x)$ is itself decreasing. $\checkmark$

Tip: "Gradient is decreasing" $\Leftrightarrow f''(x) < 0$ (concavity argument). Don't confuse with "$f$ is decreasing" which uses $f'(x) < 0$. The wording is the key: "the gradient" $\Rightarrow$ work on $f'$ as the new "function".

Question: The diagram below shows the graph of $y = f'(x)$. State the range of $x$ for which $f(x)$ is decreasing.

(Imagine: a parabola-shaped $y = f'(x)$ that crosses the $x$-axis at $x = -1$ and $x = 3$, opening upwards.)

Step 1. Reading $f'(x)$ from the graph:

• $f'(x) > 0$ (above axis): $x < -1$ or $x > 3$
• $f'(x) < 0$ (below axis): $-1 < x < 3$
• $f'(x) = 0$: $x = -1$ and $x = 3$ (these are stationary points of $f$)

Step 2. $f(x)$ is decreasing where $f'(x) < 0$.

From the graph: $f'(x) < 0$ for $-1 < x < 3$.

Answer: $f(x)$ is decreasing for $-1 < x < 3$.

Tip: When given a graph of $f'(x)$:

• Where $f'$ is ABOVE the axis $\Rightarrow$ $f$ is increasing
• Where $f'$ is BELOW the axis $\Rightarrow$ $f$ is decreasing
• Where $f'$ CROSSES the axis $\Rightarrow$ $f$ has a stationary point

DO NOT confuse the graph of $f$ with the graph of $f'$. Read the axis labels carefully.