The cross-sectional areas of a triangular prism and a right cylinder are congruent. The triangular prism has a height of 5 units
1 answer:
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The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Given:
f(x) = ln(x)
n = 4
c = 3
nth Taylor polynomial for the function, centered at c
The Taylor series for f(x) = ln x centered at 5 is:
Since, c = 5 so,
Now
f(5) = ln 5
f'(x) = 1/x ⇒ f'(5) = 1/5
f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25
f'''(x) = 2/x³ ⇒ f'''(5) = 2/5³ = 2/125
f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625
So Taylor polynomial for n = 4 is:
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Hence,
The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
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Answer:
x=-3
Step-by-step explanation:
The quadratic function determines the parabola which has the vertex at point (-3,5). The axis of symmetry passes through the vertex and is parallel to the y-axis. Hence, its equation is x=-3 (see attached diagram).
