To address the problem, let's break it down step by step.

1. Function and First-Order Taylor Approximation

Given the function $f(x) = ax^2 + bx + c$, the first-order Taylor approximation of $f(x)$ around a point $x_0$ is given by:

$T(x) = f(x_0) + f'(x_0)(x - x_0)$

Here, $f'(x_0)$ is the first derivative of $f(x)$ evaluated at $x_0$.

2. Taylor Series Expansion

The Taylor series expansion of $f(x)$ around $x_0$ is:

$f(x) = f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2}(x - x_0)^2 + \text{higher-order terms}$

The first-order Taylor approximation ignores the quadratic and higher-order terms, so:

$T(x) = f(x_0) + f'(x_0)(x - x_0)$

3. Comparison of $f(x)$ and $T(x)$

Now, let's compare the function $f(x)$ and the first-order Taylor approximation $T(x)$. The difference between the actual function and the approximation is:

$f(x) - T(x) = \frac{f''(x_0)}{2}(x - x_0)^2$

For a quadratic function $f(x) = ax^2 + bx + c$:

• $f'(x) = 2ax + b$
• $f''(x) = 2a$

So,

$f(x) - T(x) = \frac{2a}{2}(x - x_0)^2 = a(x - x_0)^2$

4. Sign of the Difference

The sign of the difference $f(x) - T(x) = a(x - x_0)^2$ depends on the coefficient $a$:

• If $a > 0$: The quadratic term $a(x - x_0)^2$ is always positive, so $f(x) > T(x)$. In this case, the first-order Taylor approximation underestimates the function $f(x)$.
• If $a < 0$: The quadratic term $a(x - x_0)^2$ is always negative, so $f(x) < T(x)$. In this case, the first-order Taylor approximation overestimates the function $f(x)$.

5. Conclusion

By examining the coefficient $a$ of the quadratic function:

• If $a > 0$, the first-order Taylor approximation is always less than $f(x)$ for any value of $x$.
• If $a < 0$, the first-order Taylor approximation is always greater than $f(x)$ for any value of $x$.

The key to deciding whether the Taylor approximation will be greater or less than $f(x)$ is simply the sign of the quadratic coefficient $a$.

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Do you want more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the second-order Taylor approximation affect the accuracy for $f(x) = ax^2 + bx + c$?
2. How does the curvature of the graph of $f(x) = ax^2 + bx + c$ relate to the sign of $a$?
3. Can you apply this concept to a cubic function? How does the Taylor series approximation change?
4. What happens to the Taylor approximation if $a = 0$?
5. How would you interpret the difference $f(x) - T(x)$ graphically?

Tip: In general, the higher the order of the Taylor approximation, the closer it is to the actual function over a wider range of $x$, especially near the point of expansion $x_0$.