Question

Sketch the graph of a function that satisfies the following conditions: f'(2) = 0, f'(0) = 1, f'(x) > 0 if 0 < x.

Answer 1

This quadratic function meets the given conditions:
`\(f'(2) = 0\), \(f'(0) = 1\)`, and
`\(f'(x) > 0\)` for
`\(0 < x\)`. So, the correct option is
`\(f(x) = (1)/(2)x^2 - 2x + 3\)`.

Explanation:

In the provided conditions, we aim to find a function that adheres to specific criteria for its derivative. The quadratic function
`\(f(x) = (1)/(2)x^2 - 2x + 3\)` is a suitable choice. Firstly, the requirement
`\(f'(2) = 0\)` suggests a possible turning point at (x = 2), characteristic of an extremum. This aligns with the quadratic nature of the function, where the coefficient of the
`\(x^2\)` term determines concavity.

Secondly, the condition
`\(f'(0) = 1\)` indicates that the slope of the tangent line at the origin is 1. This implies an upward-sloping tangent at (x = 0), consistent with the positive coefficient of the
`\(x^2\)` term, which dominates the behavior of the function for small (x).

Lastly, the requirement
`\(f'(x) > 0\) for \(0 < x\)` implies that the function is monotonically increasing for positive (x), substantiating the positive leading coefficient in the quadratic term. In essence, the selected quadratic function satisfies the given conditions by exhibiting a turning point at (x = 2), an upward slope at (x = 0), and overall positive derivatives for
`\(x > 0\)`.