Question

in problems 16-21, (a) calculate \\(f(1)\\) and (b) determine when \\(f(x) = 0\\).

1. \\(f(x) = x^2 - 5x + 13\\)
2. \\(f(x) = 5x^2 - 40x + 73\\)
3. \\(f(x) = x^3 + 9x^2 + 6\\)
4. \\(f(x) = x^3 + 3x^2 + 3x - 1\\)
5. \\(f(x) = x^3 + 2x^2 + 2x - 1\\)

Explanation:

Find the derivative of each function

Using the Polynomial Functions knowledge point, we apply the power rule for differentiation, which states that if \(g(x) = x^n\), then \(g'(x) = n x^{n-1}\).

• For 16: \(f'(x) = 2x - 5\)
• For 17: \(f'(x) = 10x - 40\)
• For 18: \(f'(x) = 3x^2 + 18x\)
• For 19: \(f'(x) = 3x^2 + 6x + 3\)
• For 20: \(f'(x) = 3x^2 + 4x + 2\)

Evaluate the derivative at x = 1

We substitute \(x = 1\) into each derived function \(f'(x)\) to find \(f'(1)\).

• For 16: \(f'(1) = 2(1) - 5 = -3\)
• For 17: \(f'(1) = 10(1) - 40 = -30\)
• For 18: \(f'(1) = 3(1)^2 + 18(1) = 21\)
• For 19: \(f'(1) = 3(1)^2 + 6(1) + 3 = 12\)
• For 20: \(f'(1) = 3(1)^2 + 4(1) + 2 = 9\)

Solve for x when the derivative is zero

We set each derivative \(f'(x) = 0\) and solve the resulting equation for \(x\).

• For 16: \(2x - 5 = 0 \implies x = \frac{5}{2}\)
• For 17: \(10x - 40 = 0 \implies x = 4\)
• For 18: \(3x^2 + 18x = 0 \implies 3x(x + 6) = 0 \implies x = 0\) or \(x = -6\)
• For 19: \(3x^2 + 6x + 3 = 0 \implies 3(x + 1)^2 = 0 \implies x = -1\)
• For 20: \(3x^2 + 4x + 2 = 0\). The discriminant is \(\Delta = 4^2 - 4(3)(2) = 16 - 24 = -8 < 0\), so there are no real solutions.

Answer:

Question 16

(a) \(f'(1) = -3\)
(b) \(f'(x) = 0\) when \(x = \frac{5}{2}\)

Question 17

(a) \(f'(1) = -30\)
(b) \(f'(x) = 0\) when \(x = 4\)

Question 18

(a) \(f'(1) = 21\)
(b) \(f'(x) = 0\) when \(x = 0, -6\)

Question 19

(a) \(f'(1) = 12\)
(b) \(f'(x) = 0\) when \(x = -1\)

Question 20

(a) \(f'(1) = 9\)
(b) \(f'(x) = 0\) has no real solutions