Question

Let $f(x) = a_{0} + a_{1}x^{2} + a_{2}x^{4} + a_{3}x^{6}, \; x \in \mathbb{R}$ where $0 < a_{0} < a_{1} < a_{2} < a_{3}$. The minimum value of $f(x)$ is

• A. $a_{0}+a_{1}+a_{2}+a_{3}$
• B. $a_{1}+a_{2}$
• C. $a_{0}$
• D. $a_{3}$
• E. $a_{0}+a_{3}$

Hint:

Math Tip: Whenever a polynomial has only even powers and all strictly positive coefficients, its graph is a "U" shape (parabola-like) that bottoms out perfectly on the y-axis. The minimum value is always just the constant term!

Answer 1

The Correct Option is C

Concept:
Calculus / Algebra - Minimum value of even functions.
Step 1: Analyze the components of the given function.
The function is $f(x) = a_0 + a_1x^2 + a_2x^4 + a_3x^6$.
Notice that the function consists entirely of even powers of $x$.
Step 2: Determine the behavior of the even powers.
For any real number $x \in \mathbb{R}$, an even power will always be non-negative: $$ x^2 \ge 0 $$ $$ x^4 \ge 0 $$ $$ x^6 \ge 0 $$
Step 3: Analyze the coefficients.
We are given the condition $0<a_0<a_1<a_2<a_3$.
This strictly means that all the coefficients ($a_1, a_2, a_3$) attached to the variable terms are positive real numbers.
Step 4: Find the minimum value of the variable terms.
Since a positive coefficient multiplied by a non-negative value is always $\ge 0$, the minimum possible value for the terms $a_1x^2$, $a_2x^4$, and $a_3x^6$ is exactly $0$.
This minimum occurs exclusively when $x = 0$.
Step 5: Evaluate the minimum value of $f(x)$.
Substitute $x = 0$ into the original function to find its global minimum: $$ f(0) = a_0 + a_1(0)^2 + a_2(0)^4 + a_3(0)^6 $$ $$ f(0) = a_0 + 0 + 0 + 0 $$ $$ f(0) = a_0 $$ Therefore, the minimum value of $f(x)$ is $a_0$.