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We are given that f(3) = 4, f'(3) = 8, f"(3) = 11. Use this information to approximate f(3.15)....

Question:

We are given that f(3) = 4, f'(3) = 8, f"(3) = 11. Use this information to approximate f(3.15). Is f increasing at x = 3? Justify your answer.

Linear Approximations:

The linear approximation of a function near a given point is the tangent line to the graph at the given point.

To linearly approximate a function {eq}\displaystyle f(x) {/eq} near {eq}\displaystyle a {/eq}, we will use the tangent line to the graph of {eq}\displaystyle f(x) {/eq} at the point {eq}\displaystyle x=a {/eq}, i.e. {eq}\displaystyle y=f(a)+f'(a)(x-a) {/eq}.

Answer and Explanation:


We know {eq}\displaystyle f(3) = 4, f'(3) = 8, f"(3) = 11 {/eq}

To approximate {eq}\displaystyle f(3.15) {/eq}, we will use the tangent line to the graph of f at {eq}\displaystyle x=3, {/eq} as show below:

{eq}\displaystyle f(x)=f(3)+f'(3)(x-3), \text{ for } x \text{ near } 3 \text{ like } x=3.15\\ \displaystyle f(3.15)\approx 4+8(3.15-3)\iff \boxed{f(3.15)\approx 5.2}. {/eq}

The function {eq}\displaystyle \boxed{\text{ is increasing at } 3, \text{ because the first derivative } f'(3)=8 \text{ is positive}}. {/eq}

The fact that the function increases at 3 is confirmed by the approximation {eq}\displaystyle f(3.15)\approx 5.2 >4=f(3). {/eq}


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Linearization of Functions

from Math 104: Calculus

Chapter 10 / Lesson 1
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