Question

In this question we will estimate the value of tan(0.163x) using the linearization of f(x)=tan(πx) at the nearby point x=1/6. a) Compute the derivative of f at 1/6. f ′(1/6)= FORMATTING: π is written pi, a is written sqrt(a). b) Write down the linearization L(x) of tan(πx) at x=1/6. L(x)= What FORMATTING: Give the exact value of all coefficients. c) Since 0.163 is near 1/6, we use the linear approximation of tan(πx) found in (b) to estimate the value of tan(πx) at x=0.163. This gives tan(0.163π)≈ FORMATTING: Round your answer to at least six decimal places. (You should check with your calculator and judge if this value is a good estimate.)

Answer 1

a) The derivative of
`\( f(x) = \tan(\pi x) \) at \( x = (1)/(6) \) is \( f'(1/6) = \pi √(3) \).`

b) The linearization
`\( L(x) \) of \( \tan(\pi x) \) at \( x = (1)/(6) \) is \( L(x) = \pi √(3)(x - (1)/(6)) + \tan((\pi)/(6)) \).`

c) Using the linear approximation,
`\( \tan(0.163\pi) \)`is approximately
`\( 0.263228 \`) (rounded to at least six decimal places).

Step-by-step explanation:

a) To find the derivative
`\( f'(x) \) of \( f(x) = \tan(\pi x) \), apply the chain rule, which gives \( f'(x) = \pi \sec^2(\pi x) \). Evaluate \( f'(1/6) \) to obtain \( f'(1/6) = \pi √(3) \).`

b) The linearization
`\( L(x) \) at \( x = (1)/(6) \) is constructed using the point-slope form of a linear equation. The equation is \( L(x) = \pi √(3)(x - (1)/(6)) + \tan((\pi)/(6)) \).`

c) Using the linear approximation,
`\( \tan(0.163\pi) \)` is estimated by substituting
`\( x = 0.163 \) into \( L(x) \)`. This yields
`\( \tan(0.163\pi) \`approx 0.263228, rounded to at least six decimal places. It's crucial to check this value with a calculator to assess the accuracy of the linear approximation. The linearization provides a good estimate when
`\( x \)` is close to
`\( (1)/(6) \)`, facilitating easier calculations than the original function.