Find the values of x... PLEASE HELP!

[Mooch22]

Let h be a function defined for all x not equal to 0 such that h(4) = -3 and the derivative of h is given by h'(x) = ((x^2)-2)/(x)) for all x not equal to 0.

a.) Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values.

b.) On what intervals, if any, is the graph of h concave up? (Justify)

c.) Write an equation for the line tangent to the graph of h at x=4. [DO YOU JUST SET THE EQUATION OF h'(x) EQUAL TO -3????]

d.) Does the line tangent to the graph of h at x=4 lie above or below the graph of h for x>4? Why?

**HELP PLEASE!!! I don't know what to do with this problem. The only thing I think I know what to do with is part c. Do you just set the derivative equal to -3? a, b, and d are foreign to me.... or maybe just the terminology...

I would appreciate the help....

 

[tkhunny]

Let h be a function defined for all x not equal to 0 such that h(4) = -3 and the derivative of h is given by h'(x) = ((x^2)-2)/(x)) for all x not equal to 0.

You COULD simply find the original function, but it isn't pretty. Note: h'(x) = x - (2/x). This gives h(x) = (1/2)x² - 2*ln(x) + C. given h(4) = -3, we have C = 2*ln(4) - 11. You may not want to do this. If you don't you may miss part a. I might have to disagree with the definition of the derivative. x ≠ 0? Maybe x > 0 would be more honest.

a.) Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values.

You have the derivative, h'(x). Where does it take on zero? Now ask yourself, does h(x) actually exist where the given h'(x) = 0?

b.) On what intervals, if any, is the graph of h concave up? (Justify)

Find the second derivative. You will see that it is always positive. What does that tell you?

c.) Write an equation for the line tangent to the graph of h at x=4. [DO YOU JUST SET THE EQUATION OF h'(x) EQUAL TO -3????]

Heard of the Point-Slope Form lately? y + 3 = h'(4)*(x - 4) You do the arithmetic.

d.) Does the line tangent to the graph of h at x=4 lie above or below the graph of h for x>4? Why?

Hmmm...This is a thought question. What is the graph doing? What would a straight line do while the graph is doing that?

OK, that's your free one. Any time someone talks like this," I don't know what to do with this problem", I become very nervous. You have to have SOME clue. You just have to.

 

[soroban]

Hello, Mooch22!

Here are the first two parts . . .

Let \(\displaystyle h(x)\) be a function defined for all \(\displaystyle x\,\neq\,0\) such that \(\displaystyle h(4)\,=\,-3\)
and the derivative is given by: \(\displaystyle h'(x)\:=\:\frac{x^2\,-\,2}{x}\) for all \(\displaystyle x\,\neq\,0\)

a.) Find all values of \(\displaystyle x\) for which the graph of \(\displaystyle h\) has a horizontal tangent,
and determine whether \(\displaystyle h\) has a local maximum, a local minimum, or neither at each of these values.

Tangents are horizontal where \(\displaystyle h'(x)\,=\,0\)

So we have: /\(\displaystyle \frac{x^2\,-\,2}{x}\:=\:0\;\;\Rightarrow\;\;x^2\,-\,2\:=\:0\;\;\Rightarrow\;\;x\,=\,\pm\sqrt{2}\) . . . (critical values)

Since \(\displaystyle h'(x)\:=\:x - 2x^{-1}\), we have: .\(\displaystyle h"(x)\:=\:1\,+\,2x^{-2}\:=\:1\,+\,\frac{2}{x^2}\)

Then: .\(\displaystyle h"(\sqrt{2})\:=\:1\,+\,\frac{2}{(\sqrt{2})^2}\:=\:1\,+\,1\;=\;2\) . . . positive, concave up: \(\displaystyle \cup\) ... minimum at \(\displaystyle x = \sqrt{2}\)

And: .\(\displaystyle h"(\)-\(\displaystyle \sqrt{2})\:=\:1\,+\,\frac{2}{(-\sqrt{2})^2}\:=\:1\,+\,1\:=\:2\) . . . positive, concave up: \(\displaystyle \cup\) ... minimum at \(\displaystyle x = -\sqrt{2}\)

b.) On what intervals, if any, is the graph of \(\displaystyle h\) concave up? (Justify)

Since \(\displaystyle h"(x)\:=\:1\,+\,\frac{2}{x^2}\) is positive for all \(\displaystyle x\,\neq\,0\),

. . the graph of \(\displaystyle h\) is concave up on: .\(\displaystyle (-\infty,0)\,\cup\,(0,\infty)\)