For x ∈ R, two real valued functions f(x) and g(x) are such that, g(x) = √x + 1 and f(g(x)) = x + 3

Q: Wait, why can't I just plug $x=0$ into the given equation $f(g(x)) = x + 3 - \sqrt{x}$?

That is a very common trap! If you plug $x=0$ into the equation, you get $f(g(0)) = 0 + 3 - \sqrt{0} = 3$. This tells you the value of $f(g(0))$, which is $f(1)$, not $f(0)$. To find $f(0)$, you need to know the rule for $f(t)$ and evaluate it at $t=0$.

Q: Do I really need to find the general expression for $f(t)$? Can't I just solve for $x$ when $g(x) = 0$?

You absolutely could! If you set $g(x) = 0$, you get $\sqrt{x} + 1 = 0$, which means $\sqrt{x} = -1$. Since the square root of a real number cannot be negative, there is no real $x$ such that $g(x) = 0$. However, the function $f(t)$ is defined for all $t$ in its domain. Finding the general expression $f(t)$ is the most robust way to handle this.

Q: Is it possible that $f(0)$ is undefined?

In this specific problem, the function $f(t)$ is a polynomial in $t$ after substitution, which is defined for all real $t$. The domain of $g(x)$ is $x \ge 0$, which means the range of $g(x)$ is $t \ge 1$. While $f(t)$ is defined for all $t$, we are only strictly 'given' the values for $t \ge 1$. However, in the context of such problems, we assume the functional form holds for the required value.

Question

For $x \in \mathbb{R}$, two real valued functions $f(x)$ and $g(x)$ are such that, $g(x) = \sqrt{x} + 1$ and $f(g(x)) = x + 3 - \sqrt{x}$. Then $f(0)$ is equal to

Options:

(A) 1

(B) 5

(C) 0

(D) $-3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are dealing with a composite function where an input $x$ passes through an inner function $g(x)$ and then an outer function $f$. We are given:

$g(x) = \sqrt{x} + 1$

$f(g(x)) = x + 3 - \sqrt{x}$

Our objective is to determine the value of $f(0)$.

### The Trap of Direct Substitution

The most common mistake is to set $x=0$ in the given equation. If you set $x=0$, you are calculating $f(g(0))$, which results in $f(\sqrt{0} + 1) = f(1)$.

This is the first lesson of composite functions: the input to the outer function $f$ is the entire output of the inner function $g$. To find $f(0)$, we must first determine the explicit rule for $f$.

### The Power of the Dummy Variable

To find the rule for $f$, we perform a transformation. We express the right-hand side of our equation, $x + 3 - \sqrt{x}$, in terms of the input to $f$, which is $g(x)$.

Let us introduce a 'dummy variable' $t$ to represent the output of the inner function:

$$t = g(x) = \sqrt{x} + 1$$

Our goal is to rewrite the expression $x + 3 - \sqrt{x}$ using only $t$.

### Building the Bridge

We have the definition $t = \sqrt{x} + 1$. To express $x$ and $\sqrt{x}$ in terms of $t$, we isolate the radical:

$$\sqrt{x} = t - 1$$

Squaring both sides allows us to solve for $x$:

$$x = (t - 1)^2$$

We have now successfully expressed both $x$ and $\sqrt{x}$ in terms of $t$.

### The Transformation

Returning to our composite function $f(g(x)) = x + 3 - \sqrt{x}$, we substitute $t = g(x)$ on the left and our expressions for $x$ and $\sqrt{x}$ on the right:

$$f(t) = (t - 1)^2 + 3 - (t - 1)$$

We have successfully transformed the function from a relationship involving $x$ into a clear, explicit rule for $f(t)$.

### The Final Step: Precision

We only need to find $f(0)$. We substitute $t = 0$ directly into our derived expression:

$$f(0) = (0 - 1)^2 + 3 - (0 - 1)$$

Simplifying the arithmetic:

$$f(0) = (-1)^2 + 3 - (-1)$$

$$f(0) = 1 + 3 + 1$$

**$f(0) = 5$**

### Conclusion

By resisting the urge to jump to conclusions and instead using the power of substitution, we have navigated the trap and arrived at the solution. The key takeaway is simple: when you see $f(g(x))$, treat $g(x)$ as a single entity, $t$.

This strategy is your most powerful tool for unlocking the secrets of composite functions. Keep practicing, keep questioning, and keep falling in love with the elegance of the math!

Animated Solution for Mathematics - Functions: For $x \in R$ , two real valued functions $f (x)$ and $g (x)$ are such that, $g (x) = x + 1$ and $f (g (x)) = x + 3 - x$ . Then $f (0)$ is equal to

Select Answer:

Visualized Solution

Understanding the Composite Function

The Substitution Strategy

Isolating the Square Root

Solving for $x$

Rewriting the Composite Function

Substituting $x$ and $x$

Targeting the Goal: $f (0)$

Simplifying the Expression

Final Arithmetic Calculation

The Sigma Insight: Composite Functions

Analyzing the Setup

The Trap of Direct Substitution

The Power of the Dummy Variable

Building the Bridge

The Transformation

The Final Step

Conclusion

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x∈R, two real valued functions f(x) and g(x) are such that, g(x)=x​+1 and f(g(x))=x+3−x​. Then f(0) is equal to