A) Use the linearization of y = ln (x) near x = 1 to estimate the value of ln (1.1). Calculate...


A) Use the linearization of {eq}y = \ln (x) {/eq} near {eq}x = 1 {/eq} to estimate the value of {eq}\ln (1.1) {/eq}. Calculate the error of this approximation with the value of {eq}\ln (1.1) {/eq} given directly by your calculator.

B) Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find the values of c guaranteed to exist by the Mean Value Theorem.

{eq}\displaystyle f (x) = x^3 - x^2 {/eq} on the interval {eq}[-1,\ 1] {/eq}.


When the function is difficult to evaluate inputs on its own, linear approximations are useful. Linear approximation is based from the Euler's method, which is denoted as {eq}y_1 = y_0 + h \cdot y_0' {/eq}, where {eq}y_0 {/eq}, {eq}h {/eq}, and {eq}y_0' {/eq}, are provided to find the {eq}y_1 {/eq} value. Linear approximation differs from Euler's method {eq}y_0 {/eq} is necessary to be determined in addition to {eq}h {/eq} value and {eq}y_{0}' {/eq}. The linearization formula is represented as {eq}f(x) = f(x_0) + h \cdot f'(x_{0}) {/eq}, where {eq}f(x)_0 {/eq}, {eq}h = x - x_0 {/eq}, and {eq}f'(x_{0}) {/eq}, are all given in order to determine the unknown {eq}f(x) {/eq} value.

Answer and Explanation:

A. Given: {eq}y = \ln(x), x = 1 {/eq}

Find: {eq}\ln(1.1) {/eq}

The point of reference will be at {eq}x = 1 {/eq}:

The step size, accordingly, will be {eq}h = 1.1-1 \Rightarrow h = 0.1 {/eq}

The derivative of {eq}y {/eq} or {eq}f(x) {/eq} in general is

{eq}\begin{align*} f'(x) &= \frac{d}{dx}(f(x)) \\ &= \frac{d}{dx}(\ln(x)) \\ &= \frac{1}{x} \\ \end{align*} {/eq}

Now all the components to compute {eq}f(1) {/eq} are found, so simply apply the linear approximation formula to find {eq}L(1.1) {/eq}.

{eq}\begin{align*} L(1.1) = f(1)+h\cdot f'(1) &= \ln(1)+(0.1)(\frac{1}{1}) \\ &= 0+(0.1)(1) \\ &= 0.1 \\ \end{align*} {/eq}

The error between the actual value and the approximated value is denoted as: {eq}\\ Error = |\frac{approximated-actual}{approximated}| {/eq}. Now the error can be determined.

{eq}\begin{align*} Error &= |\frac{0.1-\ln(1.1)}{0.1}| \\ &= |\frac{0.1-0.0953101798}{0.1}| \\ &= |\frac{0.0046898202}{0.1}| \\ &= 0.046898202 \\ \end{align*} {/eq}.

B. Given: {eq}f(x) = x^3-x^2, [-1, 1] {/eq}

We know that the given function satisfies the conditions of the Mean Value theorem because this function is a polynomial, and a polynomial function is continuous and differentiable for all real numbers.

Using the formula for the Mean Value Theorem, we need to solve for values that satisfy this theorem ({eq}c {/eq}).

{eq}\begin{align*} f'(c) = \frac{f(b)-f(a)}{b-a} &\Rightarrow \frac{d}{dc}(c^3-c^2) = \frac{f(1)-f(-1)}{1-(-1)} \\ &\Rightarrow 3\cdot c^{3-1}-2\cdot c^{2-1} = \frac{((1)^3-(1)^2)-((-1)^3-(-1)^2)}{1+1} \\ &\Rightarrow 3c^2-2c = \frac{(1-1)-(-1-1)}{2} \\ &\Rightarrow 3c^2-2c = \frac{0-(-2)}{2} \\ &\Rightarrow 3c^2-2c = \frac{0+2}{2} \\ &\Rightarrow 3c^2-2c = \frac{2}{2} \\ &\Rightarrow 3c^2-2c = 1 \\ &\Rightarrow 3c^2-2c-1 = 1-1 \\ &\Rightarrow 3c^2-2c-1 = 0 \\ &\Rightarrow (3c+1)(c-1) = 0 \\ &\Rightarrow 3c+1 = 0, c-1 = 0 \\ &\Rightarrow 3c+1-1 = 0-1, c-1+1 = 0+1 \\ &\Rightarrow 3c = -1, c = 1 \\ &\Rightarrow \frac{1}{3}\cdot 3c = \frac{1}{3}\cdot -1, c = 1 \\ &\Rightarrow c = -\frac{1}{3}, c = 1 \\ \end{align*} {/eq}

Yes, the function satisfy the hypothesis of the Mean Value Theorem at the values {eq}c = 1 {/eq} and {eq}c = -\frac{1}{3} {/eq}

Learn more about this topic:

Linearization of Functions

from Math 104: Calculus

Chapter 10 / Lesson 1

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