Gegee: f(x) = x(x - 3)^2 met f' (1) = f' (3) = 0 en f(1) = 4 8.1 Toon dat f 'n buigpunt by x = 2 het - NSC Mathematics - Question 8 - 2017 - Paper 1

To sketch the graph of f, we need to identify key points including the x-intercepts and y-intercept. We previously found that:

• The x-intercepts occur at x = 0 and x = 3 (from setting f(x) = 0).
• The y-intercept is found by evaluating f(0):

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

• The maximum occurs at (1, 4).

Plotting these points will give a rough sketch of the graph, clearly marking the points of intersection with the axes and indicating the points of inflection at x = 1.

The function $y = -f(x)$ will be concave down where the second derivative of $f(x)$ is positive, as the negative of a function flips the concavity. We previously found $f''(x) = 6x - 6$. Setting this greater than zero:

To find the local maximum of $h(x)$, we first evaluate the transformation. Since $h(x) = f(x - 2) + 3$, we shift the graph of $f(x)$ to the right by 2 units and then up by 3 units. The local maximum from $f(x)$, which occurs at (1, 4) in the original function, becomes:

$h(3) = f(1) + 3 = 4 + 3 = 7$

Thus, the local maximum coordinate in h(x) is (3, 7).

To verify Claire's assertion, we compute the first derivative at x = 2:

$f'(x) = 3x^2 - 6x + 9$

Substituting x = 2 gives: