▶️ Answer/Explanation
Solution
Step 1: Tangent Line Equation
Tangent line at \( x = a \): \( y = f(a) + f'(a)(x – a) \)
Given: \( a = 4 \), \( f(4) = -1 \), \( f'(4) = 6 \) (from graph)
Equation: \( y = -1 + 6(x – 4) \)
Step 2: Approximate f(4.5)
Evaluate at \( x = 4.5 \):
\( f(4.5) \approx -1 + 6(4.5 – 4) = -1 + 6(0.5) = -1 + 3 = 2 \)
Answer: C
▶️ Answer/Explanation
Solution
Step 1: Tangent Line Equation
Tangent line at \( x = a \): \( y = f(a) + f'(a)(x – a) \)
Given: \( a = 1 \), \( f(1) = 4 \), \( f'(x) = 2x + 1 \)
\( f'(1) = 2(1) + 1 = 3 \)
Equation: \( y = 4 + 3(x – 1) \)
Step 2: Approximate f(1.2)
Evaluate at \( x = 1.2 \):
\( f(1.2) \approx 4 + 3(1.2 – 1) = 4 + 3(0.2) = 4 + 0.6 = 4.6 \)
Answer: D
▶️ Answer/Explanation
Solution
Step 1: Tangent Line Equation
Tangent line at \( x = a \): \( y = f(a) + f'(a)(x – a) \)
Given: \( f(x) = 2 \cos x + 1 \), \( a = \frac{\pi}{2} \)
Evaluate: \( f\left(\frac{\pi}{2}\right) = 2 \cos \frac{\pi}{2} + 1 = 2(0) + 1 = 1 \)
Derivative: \( f'(x) = -2 \sin x \)
\( f’\left(\frac{\pi}{2}\right) = -2 \sin \frac{\pi}{2} = -2(1) = -2 \)
Equation: \( y = 1 – 2\left(x – \frac{\pi}{2}\right) \)
Step 2: Approximate f(1.5)
Evaluate at \( x = 1.5 \):
\( f(1.5) \approx 1 – 2\left(1.5 – \frac{\pi}{2}\right) = 1 – 2(1.5) + 2\left(\frac{\pi}{2}\right) = 1 – 3 + \pi = \pi – 2 \)
Answer: C
▶️ Answer/Explanation
Solution
Step 1: Tangent Line Equation
Tangent line at \( x = a \): \( y = f(a) + f'(a)(x – a) \)
Given: \( a = 2 \), \( f(2) = 1 \), \( f'(2) = 4 \)
Equation: \( y = 1 + 4(x – 2) \)
Step 2: Approximate f(1.9)
Evaluate at \( x = 1.9 \):
\( f(1.9) \approx 1 + 4(1.9 – 2) = 1 + 4(-0.1) = 1 – 0.4 = 0.6 \)
Answer: B