Math  /  Calculus

Question5. Consider the function h(x)=x+1xh(x)=x+\frac{1}{x}. Use the first derivative to: a) Identify (exactly) a relative maximum (x,h(x))(x, h(x)). b) Identify (exactly) a relative minimum (x,h(x))(x, h(x)).

Studdy Solution

STEP 1

What is this asking? Find the peaks and valleys of the graph of h(x)=x+1xh(x) = x + \frac{1}{x} using its derivative. Watch out! Don't forget to check if the critical points are maximums or minimums!
Also, remember to plug the *x*-values back into the original function to get the *y*-values.

STEP 2

1. Find the Derivative
2. Find Critical Points
3. Classify the Critical Points
4. Find the Maximum and Minimum

STEP 3

Alright, let's **kick things off** by finding the derivative of our function h(x)=x+1xh(x) = x + \frac{1}{x}.
Remember, the derivative tells us the *instantaneous rate of change* of our function, which is key to finding those peaks and valleys!

STEP 4

We can rewrite h(x)h(x) as h(x)=x+x1h(x) = x + x^{-1}.
This makes it easier to use the power rule!

STEP 5

Now, applying the power rule, we get h(x)=11x2h'(x) = 1 - 1 \cdot x^{-2}, which simplifies to h(x)=11x2h'(x) = 1 - \frac{1}{x^2}.
Awesome!

STEP 6

Next, we need to **find those critical points**.
Critical points are where the derivative is equal to zero or undefined.
These are the potential spots for our maximums and minimums.

STEP 7

Setting h(x)h'(x) equal to zero gives us 11x2=01 - \frac{1}{x^2} = 0.

STEP 8

Adding 1x2\frac{1}{x^2} to both sides, we get 1=1x21 = \frac{1}{x^2}.

STEP 9

Multiplying both sides by x2x^2, we have x2=1x^2 = 1.

STEP 10

Taking the square root of both sides, we find x=±1x = \pm 1.
Don't forget the plus or minus!
These are our **critical points**!

STEP 11

Now we need to **figure out** if these critical points are maximums or minimums.
We can use the second derivative test for this!

STEP 12

The second derivative of h(x)h(x) is h(x)=ddx(1x2)=2x3=2x3h''(x) = \frac{d}{dx}(1 - x^{-2}) = 2x^{-3} = \frac{2}{x^3}.

STEP 13

For x=1x = 1, h(1)=213=2h''(1) = \frac{2}{1^3} = 2, which is **positive**.
A positive second derivative means the function is concave up, so we have a **relative minimum** at x=1x = 1.

STEP 14

For x=1x = -1, h(1)=2(1)3=2h''(-1) = \frac{2}{(-1)^3} = -2, which is **negative**.
A negative second derivative means the function is concave down, so we have a **relative maximum** at x=1x = -1.

STEP 15

Almost there!
Now, we just need to plug our *x*-values back into the original function h(x)h(x) to find the *y*-values of our maximum and minimum.

STEP 16

For x=1x = 1, h(1)=1+11=2h(1) = 1 + \frac{1}{1} = 2.
So, our **relative minimum** is at (1,2)(1, 2).

STEP 17

For x=1x = -1, h(1)=1+11=2h(-1) = -1 + \frac{1}{-1} = -2.
So, our **relative maximum** is at (1,2)(-1, -2).

STEP 18

a) The relative maximum is at (1,2)(-1, -2). b) The relative minimum is at (1,2)(1, 2).

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