This Question: 1 pt 30f 11 (11 complete) a. The linear approximation to f(x)-x2
ID: 2891078 • Letter: T
Question
This Question: 1 pt 30f 11 (11 complete) a. The linear approximation to f(x)-x2 at x 0 is L(x)0 b. Linear c. If f(x) mx+ b, then the linear approximation to f at any point is Lox)-fx). d. When linear at x#0 provides a good approximation to f(x)-I a. Is statement a true or false? A. The statement is false. Substituting 0 for a in L(x)-f(a) + raxx-a) gives L(x-1. Thus, LO)-1 O B. The statement is false. Substituting 0 for a in Lox) f(a) +f(axx-a) gives L(x)-2. Thus, L(o)-2 O C. The statement is true. Substituting 0 for a in Lx)(a)+axx- a) gives L(x) -x. Thus, L(0)-0 e) D. The statement is true. Substituting O for a in L(x)-f(a)-f(a)(x-a) gives L(x)-0 Thus, Lro)-o. b. Is statement b true or false? A B. The statement is false. Linear approximation at x-o does not provide a good approximatiT to fo . x because os noit er positive nor negative. The state ment is true. Linear approximation at x-o provides a good approximation to f(x)·x) because L(X)-xfor absolute value D. The statement is true. Linear approximation at x·0 provides a good approxmaion to f )-|x) because L(x-o e. Is statement o true or faisa? absolute aluo functions. Click to select your snewer 0,3Explanation / Answer
a)
f(x)=x2
f '(x)=2x
f(0)=02=0
f '(0)=2*0=0
L(x)=f(0) +(f '(0))(x-0)
L(x)=0 +(0*(x-0))
L(x)=0
true
option D
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b)
f(x)=|x|
false
f '(x)=-1 when x<0, f '(x)=1 when x>0 , f '(x) doesnot exist for x=0
optionC
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c)
true
f(x)=mx+b
a straightline is tangent to itself
so L(x)=f(x)
option B
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d)
f(x)=lnx
f '(x)=(1/x)
L(x)=f(e) +(f '(e))(x-e)
L(x)=(lne) +(1/e)(x-e)
L(x)=1 +(x/e)-1
L(x)=(x/e)
for x=e2(which is greater than e)
f(e2)=lne2 =2lne=2*1=2
L(e2)=(e2/e)= e
we know e>2
so L(x) gives overestimte of the true values
TRUE
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