Question

question 3. consider the function ( g(x)=ln (x) ).
(a) find the tangent line approximation, ( l(x) ) at ( x = 1 ).
(b) use the tangent line approximation from (a) to estimate the value of ( ln (1.01) ).
(c) use the tangent line approximation from (a) to estimate the value of ( ln (1.3) ).
(d) of the two estimates you found in (b) and (c), which do you think is more accurate? find the error function ( e(x) ) and check your answer.
( g(1)=ln (1)=0 quad(1,0) )
( g^{prime}(x)=\frac{1}{x} )
( g^{prime}(1)=1 )
question 4. suppose it is known that for a given differentiable function ( y = g(x) ), its tangent line approximation at the point ( x=-2 ) is given by
( l(x)=-1+4(x + 2) ).
(a) compute ( l(-2) ).
(b) what must be the values of ( g(-2) ) and ( g^{prime}(-2) )?
(c) use the tangent line approximation to estimate the value of ( g(-1.95) ).

Explanation:

Step1: Compute \( L(-2) \)

Substitute \( x = -2 \) into \( L(x)=-1 + 4(x + 2) \).

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Step2: Find \( g(-2) \) and \( g'(-2) \)

The tangent line approximation formula is \( L(x)=g(a)+g'(a)(x - a) \). Here \( a=-2 \), so \( g(-2) \) is the value of \( L(x) \) at \( x = -2 \), so \( g(-2)=-1 \). And \( g'(-2) \) is the slope of the tangent line. The slope of \( L(x)=-1 + 4(x + 2) \) (in the form \( y=mx + b \)) is \( m = 4 \), so \( g'(-2)=4 \).

Step3: Estimate \( g(-1.95) \)

Substitute \( x=-1.95 \) into \( L(x)=-1 + 4(x + 2) \).

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Answer:

(a) \( L(-2)=-1 \)
(b) \( g(-2)=-1 \), \( g'(-2)=4 \)
(c) \( g(-1.95)\approx - 0.8 \)