nguyen0309
Answered

for the function f f'(x)=2x+1 and f(1)=4. What is approximation for f (1,2) found by using the line tangent to the graph of f at x=1

Answer :

LammettHash
[tex]f(1.2)\approx f(1)+f'(1)(1.2-1)=4+3\times0.2=4.6[/tex]

Compare to the actual value:

[tex]\begin{cases}f'(x)=2x+1\\f(1)=4\end{cases}\implies f(x)=x^2+x+2\implies f(1.2)=4.64[/tex]

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