Graphs of functions are given. Mark option
From the graph we observe:
Using the values: \[ f(1) = 1, \quad f(-1) = 2 \] We find a relationship: \[ f(-1) = 2f(1) \quad \Rightarrow \quad f(1) = \frac{1}{2} f(-1) \]
Suppose the function satisfies \( f(x) = 3f(-x) \). Then: \[ f(1) = 3f(-1) \quad \Rightarrow \quad 1 = 3 \times 2 = 6 \quad \text{(Incorrect)} \] Now try: \[ f(x) = \frac{1}{2} f(-x) \quad \Rightarrow \quad f(1) = \frac{1}{2} \times f(-1) = \frac{1}{2} \times 2 = 1 \quad \text{(Correct)} \] So this relation works. That means: \[ f(-x) = 2f(x) \] Which is equivalent to: \[ \boxed{f(x) = \frac{1}{2}f(-x)} \] So the best matching option among the choices is the one where: \[ f(x) = \frac{1}{2} f(-x) \] or equivalently: \[ f(-x) = 2 f(x) \]
Therefore, the correct functional relationship supported by the graph is: \[ \boxed{f(x) = \frac{1}{2} f(-x)} \quad \text{or} \quad \boxed{f(-x) = 2f(x)} \] Among multiple-choice options, this matches the option where the **negative side is scaled by 2** relative to the positive.
If the domain of the function \[ f(x)=\log\left(10x^2-17x+7\right)\left(18x^2-11x+1\right) \] is $(-\infty,a)\cup(b,c)\cup(d,\infty)-\{e\}$, then $90(a+b+c+d+e)$ equals
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: