Function problem. Need Help.

[S-Miller]

Need some assistants in find a proper method in figuring out this type of function.
The question states.

"Suppose that for some function h, h(x+5)=x²-4. Find h(3)."


I'd tried searching the web and the book that it was in (it didn't even have a answers chart for this) and other books as well. But no luck. This was my alternative option to figuring this out.
Any assistants would be appreciated.

 

[srmichael]

Need some assistants in find a proper method in figuring out this type of function.
The question states.

"Suppose that for some function h, h(x+5)=x²-4. Find h(3)."


I'd tried searching the web and the book that it was in (it didn't even have a answers chart for this) and other books as well. But no luck. This was my alternative option to figuring this out.
Any assistants would be appreciated.

What have you tried so far? Hard to help you if we don't know where you are stuck.

 

[S-Miller]

@srmichael

I tried two methods. First, by see if I move (x+5) over and seeing what formula was first used to get its answer.
Like this "h(x)=x(x+5)-5x-4=x²-4".

This was the same method I tried for this function and it got me my answer (fortunately I had the answer to this one on the last pages of the book to check.)

"Suppose that for some function g, g(x-6)=10x-1 Find g(-2)."

I move x-6 over and almost like reversed it to get its main formula "g(x)=10x+59. g(x-6)=10(x-6)+59=10x-60+59=10x-1".
So g(-2)=39. I tried this same method above.

So h(x)=x(n)-5x-4. h(3)=2x-4. I looked at this and thought this did not look right so I try using square-roots like this.

h(x+5)=x²-4. Changing h(x) to y, y=x²-4. Then moving y and x, x=y²-4. Moved the -4 over to the x, y²=x+4. And square root the x+4, y=sq-root(x+4). Got y=x+2. But this too didn't seem right as I tried x+5 to this and add right but the other method did but would confusing on a graph.

Am I on the right track or need to do something different?
Sorry for this long statement.

 

[srmichael]

@srmichael

I tried two methods. First, by see if I move (x+5) over and seeing what formula was first used to get its answer.
Like this "h(x)=x(x+5)-5x-4=x²-4".

This was the same method I tried for this function and it got me my answer (fortunately I had the answer to this one on the last pages of the book to check.)

"Suppose that for some function g, g(x-6)=10x-1 Find g(-2)."

I move x-6 over and almost like reversed it to get its main formula "g(x)=10x+59. g(x-6)=10(x-6)+59=10x-60+59=10x-1".
So g(-2)=39. I tried this same method above.

So h(x)=x(n)-5x-4. h(3)=2x-4. I looked at this and thought this did not look right so I try using square-roots like this.

h(x+5)=x²-4. Changing h(x) to y, y=x²-4. Then moving y and x, x=y²-4. Moved the -4 over to the x, y²=x+4. And square root the x+4, y=sq-root(x+4). Got y=x+2. But this too didn't seem right as I tried x+5 to this and add right but the other method did but would confusing on a graph.

Am I on the right track or need to do something different?
Sorry for this long statement.

Are you sure it is not simply asking to solve for h(3) by plugging in a value for x such that x+5 equals 3, which would mean x = -2 and thus h(3) = (-2)² - 4 = 4 - 4 = 0?

 

[HallsofIvy]

Yes, that's exactly what it is asking.

 

[S-Miller]

Are you sure it is not simply asking to solve for h(3) by plugging in a value for x such that x+5 equals 3, which would mean x = -2 and thus h(3) = (-2)² - 4 = 4 - 4 = 0?

I think it is asking what would the answer be if h(x) was 3 in place of x+5.

But the formula in your comment gave me idea and led me this conclusion.
If the function first stated is h(x+5)=x²-4 then the function to get the answer would have to be h(x)=x²-29 if it was x.
So with that h(x+5)=x²-29=(x+5)²-29=x²+25-29=x²-4.
So with figured out, if h(3) it would be h(3)=x²-29=3²-29=9-29

So I guess it is asking if h(x+5)=x²-4, then what would h(3) be, which would be h(3)=-20.

Please tell if this makes any sense to anyone.

 

[S-Miller]

In retrospect I didn't think about squaring the number x and tried to times(multiple) x with x.

 

[HallsofIvy]

I think it is asking what would the answer be if h(x) was 3 in place of x+5.

No, you were right before.
h(3) will be the same as h(x+5) if x+ 5= 3. And that tells you that x= -2.
Since \(\displaystyle h(x+ 5)= x^2- 4\), it follows that \(\displaystyle h(3)= h(-2+ 5)= (-2)^2- 4= 0\).

But the formula in your comment gave me idea and led me this conclusion.
If the function first stated is h(x+5)=x²-4 then the function to get the answer would have to be h(x)=x²-29 if it was x.

I'm sorry, you will have to explain how you got that! From \(\displaystyle h(x+5)= x^2- 4\), we can say, if y= x+ 5, then x= y- 5 so that \(\displaystyle h(x+5)= h(y)= (y- 5)^2- 4= y^2- 10y+ 25- 4= y^2- 10y+ 21\). That is, of course, the same as saying that \(\displaystyle h(x)= x^2- 10x+ 2\). From that \(\displaystyle h(3)= 3^2- 10(3)+ 21= 9- 30+ 21= 0\) again.

So with that h(x+5)=x²-29=(x+5)²-29=x²+25-29=x²-4.
So with figured out, if h(3) it would be h(3)=x²-29=3²-29=9-29

So I guess it is asking if h(x+5)=x²-4, then what would h(3) be, which would be h(3)=-20.

Please tell if this makes any sense to anyone.

 

[JeffM]

Need some assistants in find a proper method in figuring out this type of function.
The question states.

"Suppose that for some function h, h(x+5)=x²-4. Find h(3)."


I'd tried searching the web and the book that it was in (it didn't even have a answers chart for this) and other books as well. But no luck. This was my alternative option to figuring this out.
Any assistants would be appreciated.

Use the method of substitution for problems of this type.

\(\displaystyle Let\ y = x + 5 \implies x = y - 5 \implies h(x + 5) = h(y) = x^2 - 4 = (y - 5)^2 - 4 = y^2 -10y + 21.\)

Now the problem is straight forward.

\(\displaystyle y = 3 \implies h(y) = h(3) = 3^2 - 10 * 3 + 21 = 9 - 30 + 21 = 0.\)

EDIT: My method and Hall's method are logically identical. Use whichever you personally find most intuitive. They will get you to the same place.

 

[S-Miller]

I'm sorry, you will have to explain how you got that!

Use the method of substitution for problems of this type.

\(\displaystyle Let\ y = x + 5 \implies x = y - 5 \implies h(x + 5) = h(y) = x^2 - 4 = (y - 5)^2 - 4 = y^2 -10y + 21.\)

Now the problem is straight forward.

\(\displaystyle y = 3 \implies h(y) = h(3) = 3^2 - 10 * 3 + 21 = 9 - 30 + 21 = 0.\)

EDIT: My method and Hall's method are logically identical. Use whichever you personally find most intuitive. They will get you to the same place.

I don't think that might be it. Don't mean any disrespect to ether you or your math knowledge. I tried to reverse y and x and try the function and didn't get the same results. I might not have been clear as I need to be, so that fault falls on me.

But let me use this function again (which stated, had the answer at the end of the book)
"Suppose for some function g, g(x-6)=10x-1. Find g(-2)."
I just thought up a function that would result in the output of 10x-1 for g(x) if x-6 was x, so...

\(\displaystyle g(x-6) = 10 x + 59 = 10 (x-6) + 59 = 10x - 1\)

So, with the function being g(x)=10x+59 I did the same function with the value -2.

\(\displaystyle x = -2 \implies g(-2) = 10 x + 59 = 10 (-2) +59 = 39 \)

With g(-2)=39 (which the book says it is the correct answer, unfortunately it didn't give for the function previously asked.)
I applied this method to the function on topic coming to somewhat like a writers block with math, to this.

So with h(x+5)=x²-4, if the value was x what would its output be. Which would be...

\(\displaystyle h(x+5) = x^2 - 29 = (x + 5)^2 - 29 = x^2 + 25 -29 = x^2 - 4\)

So i did this with 3.

\(\displaystyle x = 3 \implies h(3) = x^2 - 29 = (3)^2 -29 = 9 -29 = -20\)


@HallsofIvy
Hope this explains it for you and how I got this result.

So, as with the substitution method ether I must have not explained the problem correctly or I must be doing it wrong to not get my answer. Sorry.
Either way, I do appreciated the advice and help everyone is giving.

 

[MarkFL]

HallsofIvy and JeffM have gotten the correct result. Although their methods are shorter and (much) more straightforward, here's another approach:

\(\displaystyle h(x+5)=x^2-4=x^2+10x+25-10x-50+21=(x+5)^2-10(x+5)+21\) hence:

\(\displaystyle h(3)=3^2-10(3)+21=0\)

 

[S-Miller]

HallsofIvy and JeffM have gotten the correct result. Although their methods are shorter and (much) more straightforward, here's another approach:

\(\displaystyle h(x+5)=x^2-4=x^2+10x+25-10x-50+21=(x+5)^2-10(x+5)+21\) hence:

\(\displaystyle h(3)=3^2-10(3)+21=0\)

Holy ****. Did the function on what you showed and it fills out correctly, I wonder what I keep doing wrong. I guess I've been working it wrong with squaring (x+5) I believe. You seem to made it clear now. Thanks.

Now I can sleep a little easier knowing that it is figured out.
Thanks again MarkFL for your time and comment and thanks JeffM, HallsofIvy and srmichael.