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Use local linearization to find the estimate of f(-0.9), given f(x) = 2x^3 + 5x.

Question:

Use local linearization to find the estimate of {eq}f(-0.9) {/eq}, given {eq}f(x) = 2x^3 + 5x {/eq}.

How to Estimate Function Values Using Linearization:

We have to estimate of f(-0.9) by using local linearization. We will first find the value of f(x) and f'(x) at x=1, then use the formula of linearization and substitute the values to get the desired result.

Answer and Explanation:

The given function is {eq}f(x) = 2x^3 + 5x {/eq}.

First find the function at {eq}x=1, {/eq} then the function will be $$\begin{align*} f\left( x \right) = 2{x^3} + 5x\\ f\left( 1 \right) &= 2{\left( 1 \right)^3} + 5\left( 1 \right)\\ f\left( 1 \right) &= 2 + 5\\ f\left( 1 \right) &= 7. \end{align*} $$

Take the derivative of the function with respect to "x" and we have $$\begin{align*} f\left( x \right) &= 2{x^3} + 5x\\ \frac{d}{{dx}}\left[ {f\left( x \right)} \right] &= \frac{d}{{dx}}\left[ {2{x^3} + 5x} \right]\\ f'\left( x \right) &= 6{x^2} + 5. \end{align*} $$

At {eq}x=1, {/eq} then the function will be $$\begin{align*} f'\left( x \right) &= 6{x^2} + 5\\ f'\left( 1 \right) &= 6{\left( 1 \right)^2} + 5\\ f'\left( 1 \right) &= 11. \end{align*} $$

Use local linearization and we have $$\begin{align*} v(t)&=h'(t)\\ &=\frac{d}{dt}\left(128t - 18t^2 \right)\\ &=128-18\cdot 2t\\ &=128-36t\\ \end{align*} $$

At {eq}x=-0.9, {/eq} then the value of function will be $$\begin{align*} f\left( x \right) &\approx f\left( a \right) + f'\left( a \right)\left( {x - a} \right)\\ f\left( x \right) &\approx f\left( 1 \right) + f'\left( 1 \right)\left( {x - 1} \right)\\ f\left( { - 0.9} \right) &\approx f\left( 1 \right) + f'\left( 1 \right)\left( { - 0.9 - 1} \right)\\ &\approx 7 + 11\left( { - 1.9} \right)\\ &\approx 7 - 20.9\\ f\left( { - 0.9} \right) &\approx - 13.9. \end{align*} $$


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How to Estimate Function Values Using Linearization

from Math 104: Calculus

Chapter 10 / Lesson 2
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