Question

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 1

`f(1.2)\approx f(1)+f'(1)(1.2-1)=4+3*0.2=4.6`

Compare to the actual value:


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