e3r2a1cakCh7

Answered

2022-12-01

Let f be a differentiable function such that f(3) = 2 and f'(3) = 5. If the tangent line to the graph of f at x = 3 is used to find an approximation to a zero of f, that approximation is?

Answer & Explanation

ysik92ASw

Expert

2022-12-02Added 13 answers

Find an approximate value of a zero of f(x) near a=3. That is to say, you're looking for some value c such that f(c)=0, but all you have at your disposal is the linear approximation to the function.

$f(x)\approx f(3)+{f}^{\prime}(3)(x-3)\phantom{\rule{0ex}{0ex}}f(x)\approx 2+5(x-2)\phantom{\rule{0ex}{0ex}}f(x)\approx 5x-8$

You know that if c is a zero of , f(c)=0, so you get

$f(c)\approx 5c-8\phantom{\rule{0ex}{0ex}}0=5c-8\Rightarrow 5c=8\Rightarrow c=\frac{8}{5}=1.6$An approximate zero of f(x) is x=1.6.

$f(x)\approx f(3)+{f}^{\prime}(3)(x-3)\phantom{\rule{0ex}{0ex}}f(x)\approx 2+5(x-2)\phantom{\rule{0ex}{0ex}}f(x)\approx 5x-8$

You know that if c is a zero of , f(c)=0, so you get

$f(c)\approx 5c-8\phantom{\rule{0ex}{0ex}}0=5c-8\Rightarrow 5c=8\Rightarrow c=\frac{8}{5}=1.6$An approximate zero of f(x) is x=1.6.

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