Math help?

[Yusuke Urameshi]

Calculus help?

I posted a problem yesterday that I needed help with. I was given the go-ahead to post any other problems that I can't figure out, so here it is:

y = 0.5 ·4^x

the equation of the line being found is tangent to the graph at = (0, 0.5)

I'm supposed to find the x-value for where the tangent line crosses the x-axis. Any help is appreciated and rep will be given for the correct answer or legitimate attempts for the correct answer. Thanks, guys.

 

[Zee Seh]

Re: Calculus help?

Need more explanation.. Which graph?

 

[Zantos]

Re: Calculus help?

it never crosses the x-axis ! just put y=0 and you will see there is no amount for x that makes y=0 !! there is how ever a convergence line ( horizontal ) and that's when x goes to minus infinity !

 

[Zee Seh]

Re: Calculus help?

Well if i havent misinterpreted the question. The x value should be 2.51 approx.

 

[Rokudaime Hokage Naruto]

Re: Calculus help?

1) Differentiate y = 0.5 ·4^x
2) Substitute the given X value 0.5 into the differential you just calculated - this will give you the gradient of your tangent (lets call it 'm')
3) Use the identity y=mx + c, substituting in your values of y, m and x in order to find c.
4) Make y=0 in this equation and rearrange to find X.

Sorry I can't be bothered to calculate it myself but this should help you.

 

[Rainbow Dash]

Re: Calculus help?

If you get stuck on problems like this, the first thing you should do is graph the equation on your graphing calculator (or on Google) so you see a visual of what you're trying to get.

 

[Zee Seh]

Re: Calculus help?

x= -1/log(0.4) . . . Is the only possible answer ( ._.)

 

[fiend]

y=0.5*4^x, differentiates to y'= 4^x ln(2)

Now, value of slope/gradient of tangent at (0,0.5) is y'=(4^0)*ln(2) which means slope of the equation of tangent is ln(2)

Equation of tangent at (0,0.5) becomes (y-0.5)=ln(2)x

It intersects with x axis at y=0

⇒ 0-0.5= ln(2)x

which means x=-1/2*ln(2)

or x=-1/ln(4)

⇒x=0.72134

where ln= natural log

I hope, I am correct doing these after many months.

 

[Agent Phrank]

If the question is to find such line tangent to that function at (0, 0.5) then it's pretty easy;
first find the first derivative to y
y'=0.5((4^x))(ln(4)))=0.6931(4^x)

then evaluate that at x=0, which gives .6931 this is the slope of the tangent line.

Now, if we evaluate x=0 in the original function, we know y=0.5 there, so now that we have the slope and a point, we can find the equation for the tangent line.

y=mx+b
where m=0.6931 and b=0.5

so, its y=0.6931 x +0.5

 

[Zee Seh]

y=0.5*4^x

That isnt .4? Fack. I was supposing it to be 0.4 this whole tym ( ._.)

Answer is absolutely correct mate

 

[fiend]

That isnt .4? Fack. I was supposing it to be 0.4 this whole tym ( ._.)

Answer is absolutely correct mate

The "." creates many problems that's why * is better.

Since you were thinking it to be 0.4 so your answer is also correct.

 

[Mille Ignis]

Re: Calculus help?

I know this doesn't belong on this site, but why not ask, right? If any of you guys know how to solve this, help would be greatly appreciated. The prof. never taught us how to do this. If you could explain how you got what you did so I can understand it in the future, that'd be awesome. Thanks, guys!

2. If f(x)=ax^2+b, f(−5)=6 and f′(−5)=2, find the coefficients a and b.

This one is fairly simple.

Since f(x)=ax^2 + b then f'(x)=2ax since b is a constant, the derivate of a constant is zero

We know that f'(-5) = 2 so,
2 = 2a(-5)
Sove for a, and you should get
a = -1/5

Next to solve for b, we use the first equation.

f(x) = ax^2 + b = (-1/5)x^2 + b

since f(-5) = 6 = (-1/5)(-5)^2 + b

Solve for b and you should get

b = 11

So, a = -1/5 and b = 11

I posted a problem yesterday that I needed help with. I was given the go-ahead to post any other problems that I can't figure out, so here it is:

y = 0.5 ·4^x

the equation of the line being found is tangent to the graph at = (0, 0.5)

I'm supposed to find the x-value for where the tangent line crosses the x-axis. Any help is appreciated and rep will be given for the correct answer or legitimate attempts for the correct answer. Thanks, guys.

We know that to get the equation for the tangent line of the function, we must take the derivative.
f(x) = 0.5 * 4^x

This derivative is a little tricky.

First lets get x out of the exponent by using the natural log, so
ln[f(x)] = ln[0.5] + xln[4] Recall that the ln[ab] = ln[a] + ln(b)

so now we can take the derivative.

[1/f(x)]f'(x) = ln[4]

thus, f'(x) = ln[4]f(x) = ln4[(0.5)*4^x]

Now that we have the equation of the tangent line, we want to know the slope of the line at x = 0 Taken from the coordinate (0,0.5)
So f'(0) = ln[4][(0.5)(4^x)] = (0.5)*ln[4]

Then, using the equation y2-y1 = m(x2-x1) we can find the equation of the tangent line that passes through the point (0,0.5)

so plugging in our numbers we get

y2 - 0.5 = (0.5)*ln[4](x2-0)

Thus,

y2 = (0.5)*ln[4]x2 + 0.5

Now to find the x value for the tangent line when it crosses the x-axis we let y=0
So,

0 = (0.5)*ln[4]x + 0.5

x = [-(0.5)/(0.5)ln4] = -1/ln[4]

I hope this helps. Sorry if I don't do a good job of explaining what I did and why.

 

[Uzumaki Naruta]

I know this doesn't belong on this site, but why not ask, right? If any of you guys know how to solve this, help would be greatly appreciated. The prof. never taught us how to do this. If you could explain how you got what you did so I can understand it in the future, that'd be awesome. Thanks, guys!

2. If f(x)=ax^2+b, f(−5)=6 and f′(−5)=2, find the coefficients a and b.

f´(x) = 2ax

f´(-5) = 2*a*(-5) = 2
-10a = 2
a = -1/5

f(x) = (-1/5)x^2 + b

f(-5) = (-1/5)*(-5)^2 + b = 6
(-1/5)*25 + b = 6
-5 + b = 6
b = 11