Question:

Let $f(x) = \sqrt[3]{x^2 + 4}$. Consider the point $P(3, f(3))$ on the graph of $f$.

a) Use linear approximation to estimate the value of $f(3.02)$.

b) Determine an upper bound for the absolute error in the linear approximation obtained in part (a).

c) Find the differential $dy$ and evaluate $dy$ when $x = 3$ and $dx = 0.02$.

Answer:

a) Estimating $f(3.02)$ using linear approximation:

The linear approximation of a function $f$ at a point $(a, f(a))$ can be expressed as:

$$L(x) = f(a) + f'(a)(x - a)$$

where $f'(a)$ is the derivative of $f$ evaluated at $a$. In this case, $a=3$ and $f(a) = f(3)$.

To find $f'(3)$, we differentiate $f(x)$ with respect to $x$, which gives:

$$f'(x) = \frac{2}{3 \sqrt[3]{x^2 + 4}} \cdot \frac{1}{2}x$$

Thus, $f'(3) = \frac{2}{3 \sqrt[3]{3^2 + 4}} \cdot \frac{1}{2} \cdot 3$ becomes:

$$f'(3) = \frac{3}{4 \sqrt[3]{7}}$$

Now, using the linear approximation formula, we have:

$$L(x) = f(3) + f'(3) \cdot (x - 3)$$

Substituting the values, we get:

$$L(x) = \sqrt[3]{3^2 + 4} + \frac{3}{4 \sqrt[3]{7}} \cdot (x - 3)$$

For $x = 3.02$, the estimate for $f(3.02)$ is:

$$L(3.02) = \sqrt[3]{3^2 + 4} + \frac{3}{4 \sqrt[3]{7}} \cdot (3.02 - 3)$$

Simplifying the above expression gives:

$$L(3.02) = \sqrt[3]{13} + \frac{3}{4 \sqrt[3]{7}} \cdot 0.02$$

Finally, the estimate for $f(3.02)$ is approximately:

$$L(3.02) \approx \sqrt[3]{13} + \frac{0.03}{4 \sqrt[3]{7}}$$

b) Upper bound for the absolute error:

The upper bound for the absolute error in linear approximation can be determined using the formula:

$$E(x) = \frac{f''(c)}{2}(x-a)^2$$

where $f''(c)$ is the second derivative of $f$ evaluated at some point $c$ in the interval between $a$ and $x$.

To find the upper bound, we need to find the maximum value of $f''(x)$ in the interval [3, 3.02].

Differentiating $f'(x)$ with respect to $x$ gives:

$$f''(x) = -\frac{8x}{9(x^2 + 4)^{\frac{5}{3}}}$$

Finding the critical points by solving $f''(x) = 0$:

$$-\frac{8x}{9(x^2 + 4)^{\frac{5}{3}}} = 0$$

This equation holds true when $x = 0$.

Evaluating $f''(x)$ at the endpoints and critical points, we have:

$$f''(3) = -\frac{8(3)}{9(3^2 + 4)^{\frac{5}{3}}}$$

$$f''(0) = -\frac{8(0)}{9(0^2 + 4)^{\frac{5}{3}}}$$

Calculating these values yields:

$$f''(3) = -\frac{8}{81\sqrt[3]{7}}$$

$$f''(0) = 0$$

Since $f''(0)$ is not defined, we consider $f''(c) = f''(3)$ as the maximum value for the second derivative.

Substituting these values into the upper bound formula, we get:

$$E(3.02) = \frac{f''(3)}{2}(3.02 - 3)^2$$

Simplifying the expression gives:

$$E(3.02) = \frac{-\frac{8}{81\sqrt[3]{7}}}{2}(0.02)^2$$

Finally, the upper bound for the absolute error in the linear approximation is approximately:

$$E(3.02) \approx \frac{-1}{810\sqrt[3]{7}}$$

c) Finding the differential $dy$ and evaluating $dy$ when $x = 3$ and $dx = 0.02$:

The differential $dy$ represents the change in $y$ (or $f(x)$) due to a small change in $x$, denoted as $dx$. It can be calculated using the formula:

$$dy = f'(a) \cdot dx$$

where $a$ is the value of $x$ and $dx$ represents the change in $x$.

Given that $a = 3$ and $dx = 0.02$, we already calculated $f'(3)$ as:

$$f'(3) = \frac{3}{4 \sqrt[3]{7}}$$

Substituting these values, we have:

$$dy = \frac{3}{4 \sqrt[3]{7}} \cdot 0.02$$

Evaluating the expression, we get:

$$dy \approx \frac{0.03}{2 \sqrt[3]{7}}$$

Thus, $dy$ when $x = 3$ and $dx = 0.02$ is approximately:

$$dy \approx \frac{0.03}{2 \sqrt[3]{7}}$$