How to Find f(x) Given f(x^3) 5x^2

When faced with a problem of the form f(x^3) 5x^2 and you are asked to find f(x), it is advantageous to use algebraic substitution to simplify and solve the problem. This article will guide you through finding f(x) using several methods, all leading to the same answer: f(x) 5x - 13.

Method 1: Using y x^3

Let's begin by setting y x^3. By substituting y for x^3, we can express x in terms of y, thereby simplifying the given equation. Since y x^3, we have:

y x^3
Canceling the cube operation, we take the cube root of both sides:
x y^{1/3}

Now, substitute x with y^{1/3} in the original equation:

f(y) 5(y^{1/3})^2 5y^{2/3}

We now need to express this in terms of x. Recall that y x^3, so:

5y^{2/3} 5(x^3)^{2/3} 5x^2

We can now generalize it as:

5x^2 - 13

Since y is just a dummy variable, we can replace y back with x:

f(x) 5x - 13

Method 2: Substitution and Simplification

In this method, we will use the substitution x → x - 3. Let's start with the original equation:

f(x^3) 5x^2

If we let x - 3 t, then x t 3. Now, let's substitute x with t 3 in the equation:

Replacing x with t 3 gives:

f((t 3)^3) 5(t 3)^2

However, this approach may seem more complex. Instead, let's stick with the simpler substitution:

Focusing on a direct substitution, we let x t, then x-3 t-3. Therefore:

f(t-3^3) 5(t-3^2)

Which simplifies to:

f(t) 5t - 15^2

And finally:

f(t) 5t - 13

Replacing t with x, we have:

f(x) 5x - 13

Method 3: Direct Substitution and Simplification

Another straightforward approach is to let t x^3. By doing so, we can directly manipulate the equation:

Taking t x^3 and thus:

t x^3
x t^{1/3}

Substituting x with t in the original equation:

f(t) 5(t^{1/3})^2 5t^{2/3}

Since t x^3, we have:

5t^2 - 13

Replacing t with x, we get:

f(x) 5x - 13

Conclusion

To summarize, the given function f(x^3) 5x^2 can be transformed into the function f(x) 5x - 13 using various substitution and simplification techniques. Whether through direct substitution, algebraic manipulation, or a combination of both, the result remains consistent. This demonstrates the power of algebraic methods in unlocking the solution to complex polynomial functions.

Understanding these methods can be beneficial for students and professionals in mathematics, engineering, and related fields where polynomial functions play a crucial role.