Q.Find the equations of the tangent and the normal, to the curve $16x^2 + 9y^2 = 145$ at the point $(x_1, y_1)$, where $x_1 = 2$ and $y_1 > 0$. OR Find the intervals in which the function $f(x) = \dfrac{x^4}{4} - x^3 - 5x^2 + 24x + 12$ is

The Intuition: What Does "Increasing" Really Mean?

Imagine walking along the graph of a function from left to right. If the function is increasing, then as you step right (increasing $x$), you always move upward — your height $f(x)$ never drops. You might stay flat briefly, but you never go down.

That's the visual idea. But we need a precise way to check it without drawing the entire graph — that's the Increasing Function Test, which uses the derivative to tell you where a function is rising.

The Core Idea: Derivative as a Slope Detector

The derivative $f'(x)$ gives the slope of the tangent line — the instantaneous rate of change. Positive slope means the function is rising at that instant; negative means falling. So the natural question: if the derivative is positive everywhere on an interval, does that guarantee the function is increasing on that whole interval? The answer is yes — and that's the Increasing Function Test.

The Precise Statement

The conditions "continuous on the closed interval" and "differentiable on the open interval" ensure there are no jumps or corners that could break the logic.

Why Does This Work? (A Quick Proof Sketch)

The proof relies on the Mean Value Theorem. For $x_1 < x_2$ in $[a, b]$, there exists some $c$ between them such that:

Since $x_2 - x_1 > 0$, if $f'(c) > 0$ the right-hand side is positive, so $f(x_2) > f(x_1)$. This holds for any pair $x_1 < x_2$ — exactly the definition of strictly increasing.

How to Use It in Practice

1. Compute $f'(x)$.
2. Solve $f'(x) > 0$ — the solution intervals tell you where $f$ is strictly increasing.
3. Check endpoints if needed. …