Q.Prove that the following functions do not have maxima or minima:

Concept: Monotonic Function Analysis – a strictly monotonic function on its domain cannot have interior local extrema.

(i) $f(x) = e^x$

$f'(x) = e^x > 0$ for all $x \in \mathbb{R}$. The function is strictly increasing on its entire domain. No turning points exist, so no maxima or minima.

(ii) $g(x) = \log x$

$g'(x) = \frac{1}{x} > 0$ for all $x > 0$. Strictly increasing on $(0, \infty)$. No critical points, hence no extrema.

(iii) $h(x) = x^3 + x^2 + x + 1$ …

We need to show that each of these functions has no local maxima or minima anywhere on its domain. The standard approach is to examine the derivative and see whether it ever equals zero — because a necessary condition for a local extremum (at an interior point) is that the derivative is zero there.

So if we can show that $f'(x) \neq 0$ for every $x$ in the domain, then no local extremum can exist. Let's check each function.

1. $f(x) = e^x$

Domain: All real numbers $\mathbb{R}$.

Derivative: $f'(x) = e^x$.

Now, $e^x > 0$ for every real $x$. It is never zero. There is simply no $x$ such that $e^x = 0$.

Since the derivative never vanishes, there is no critical point. Therefore, $f(x) = e^x$ has no local maximum or minimum.

2. $g(x) = \log x$

Domain: $(0, \infty)$ (positive real numbers).

Derivative: $g'(x) = \frac{1}{x}$.

For $x > 0$, $\frac{1}{x} > 0$. It is never zero. Again, no critical point exists.

Thus $g(x) = \log x$ has no local maximum or minimum.

--- …

Method: Proving a Function Has No Local Extrema

This method proves the absence of local maxima/minima, which is a different task from finding one — the goal is to show the necessary condition for an extremum can never be satisfied.

Steps

Step 1: Recall the necessary condition for a local extremum.

If a differentiable function $f$ has a local maximum or minimum at an interior point $c$, then $f'(c) = 0$ (Fermat's theorem). This is a necessary condition, not a sufficient one — but it is exactly what we use in reverse here.

Step 2: Compute $f'(x)$ and show it can never equal zero anywhere in the domain.

• For an exponential-type function, use the fact that an exponential is always strictly positive.
• For a logarithmic-type function, use the fact that $\frac{1}{x} > 0$ throughout the domain $x>0$.
• For a polynomial whose derivative is itself a quadratic, compute the discriminant $\Delta = b^2-4ac$ of that derivative.

Step 3: Interpret a negative discriminant as "no real solution". …

Common Mistakes

Mistake 1: Concluding $g(x)=\log x$ has an extremum at $x=1$ because $\log 1 = 0$.

Why it's wrong: a zero function value has nothing to do with a zero derivative — the actual derivative there is $g'(1)=\frac11=1\neq 0$, so $x=1$ is not even a critical point. Correct approach: only check where $f'(x)=0$, never where $f(x)=0$, when hunting for extrema.

Mistake 2: Stopping at "the derivative is a quadratic" without checking whether it actually has real roots.

Why it's wrong: for $h(x)=x^3+x^2+x+1$, a student who solves $3x^2+2x+1=0$ using the quadratic formula without first checking the discriminant may report spurious complex "critical points" as if they were real. Correct approach: compute the discriminant first — a negative value ($\Delta=-8$ here) proves immediately that no real critical point exists, without needing the formula at all. …