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Q5. Find all the relative extreme points and inflection points of $y=x^{3}-12 x...

Sep 19, 2024

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Solution by Steps

For Q5: Find all the relative extreme points and inflection points of $y=x^{3}-12x+10$.

step 1

To find the relative extreme points, we first find the derivative of the function:

y' = 3x^2 - 12

step 2

Setting the derivative equal to zero to find critical points:

3x^2 - 12 = 0

step 3

Solving for

x

x^2 = 4 \Rightarrow x = ±2

step 4

To determine if these points are relative extrema, we check the second derivative:

y'' = 6x

step 5

Evaluating the second derivative at the critical points: y''(2) = 12 > 0 (local minimum) and y''(-2) = -12 < 0 (local maximum)

step 6

The relative extreme points are:

(-2, y(-2))

and

(2, y(2))

. Calculating

y(-2) = 10 + 24 - 12 = 22

and

y(2) = 10 - 24 + 12 = -2

step 7

To find inflection points, we set the second derivative equal to zero:

6x = 0 \Rightarrow x = 0

step 8

The inflection point is at

(0, y(0)) = (0, 10)

Answer

Relative extreme points:

(-2, 22)

(local maximum) and

(2, -2)

(local minimum); Inflection point:

(0, 10)

--- For Q6: Find the linear approximation of the function $f(x)=\sqrt{x+9}$ at $x=0$ and use it to approximate the number $\sqrt{9.01}$.

step 1

The linear approximation formula is given by:

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

, where

a = 0

step 2

First, we find

f(0) = \sqrt{0 + 9} = 3

step 3

Next, we find the derivative:

f'(x) = \frac{1}{2\sqrt{x + 9}}

. Evaluating at

x = 0

f'(0) = \frac{1}{2\sqrt{9}} = \frac{1}{6}

step 4

Now we can write the linear approximation:

L(x) = 3 + \frac{1}{6}(x - 0) = 3 + \frac{1}{6}x

step 5

To approximate

\sqrt{9.01}

, we substitute

x = 0.01

L(0.01) = 3 + \frac{1}{6}(0.01) = 3 + \frac{0.01}{6} = 3 + 0.0016667 \approx 3.00167

step 6

Rounding to five decimal places, we get:

3.00167

Answer

The linear approximation of

\sqrt{9.01}

is approximately

3.00167

Key Concept

Finding relative extrema and inflection points involves using derivatives to analyze the behavior of functions. Linear approximation uses derivatives to estimate function values near a point.

Explanation

The relative extreme points and inflection points were determined using first and second derivatives, while the linear approximation provided a method to estimate values of functions close to a known point.