How Do I Find the Local Max of My Function?

[JustHere155]

Homework Statement

Homework Equations

Fundamental Theorem of Calculus

The Attempt at a Solution

I tried to find the max/min value for x^2+15x-36 and got x=-3 or x-12. I don't know which one I need to use to get the Local max of my function >.<.

 

[Mark44]

Homework Statement

Homework Equations

Fundamental Theorem of Calculus

The Attempt at a Solution

I tried to find the max/min value for x^2+15x-36 and got x=-3 or x-12. I don't know which one I need to use to get the Local max of my function >.<.

How do you usually find the maximum or minimum values for a function? Doesn't it involve taking the derivative and setting it to 0?

This problem looks like an application of the Fundamental Theorem of Calculus, so read up on that in your textbook.

 

[JustHere155]

Ok here goes nothing,

I know from FTC, that g(x)=f'(x)
So in my case replace all the x with t and make f(t)= 0 to find the critical points
I get ((t²+15t+36)/(1+cos²(t)))=0 I found t=-12/-3
Next I take the f"(t) using the quotient rule to find my local max. I get:
((2t+15)(1+cos^2(t))-(t^2+15t+36)(2sin(t)cos(t)))/(1+cos^2(t))^2 --> simplified gives me
(2t+15)/(1+cos^2(t)) at this point I plug my -3 and -12 and I respectively get: 4.54/-5.25
so my local max is (-3,4.54)

This is a long a shot. I hope this is right. Can anyone put some feed back into this?

 

[jackmell]

I don't see why you're looking for the extremum if you're looking for t. Suppose you have:

u=\int_0^t x^2dx

then:

u=t^3/3=g(t)

we then take the inverse of g ang obtain:

t=g^{-1}(u)=\sqrt[3]{3u}

so I think it's the same here:

u=\int_0^t \frac{x^2+15x+36}{1+\cos^2(x)}dx=g(t)

and then:

t=g^{-1}(u)

now, we can't directly invert g in this case but the integral is one-to-one so it has a unique inverse. So you said find "t" right? Well to me that means any way so I'd (numerically) create a table of {t,u(t)}, make a least square fit of that data, then bam-o. Got t.

Hope I'm not being out of line here. Just trying to help.

 

[Mark44]

Ok here goes nothing,

I know from FTC, that g(x)=f'(x)
So in my case replace all the x with t and make f(t)= 0 to find the critical points
I get ((t²+15t+36)/(1+cos²(t)))=0 I found t=-12/-3

Better to write this as t = -12 or t = -3. The way you had it, I thought you meant a single number, -12/-3, which is 4.

Next I take the f"(t) using the quotient rule to find my local max. I get:
((2t+15)(1+cos^2(t))-(t^2+15t+36)(2sin(t)cos(t)))/(1+cos^2(t))^2 --> simplified gives me
(2t+15)/(1+cos^2(t)) at this point I plug my -3 and -12 and I respectively get: 4.54/-5.25
so my local max is (-3,4.54)

This is a long a shot. I hope this is right. Can anyone put some feed back into this?

I didn't check your work, so it might or might not be right. There's another approach, which doesn't require the use of the second derivative. At each criticlal point, if the first derivative changes sign, the critical point is a local maximum or local minimum.

That's pretty easy with this problem, where f'(t) = (t² + 15t + 36)/(1 + cos²(t)).

The denominator is always > 0, so all you have to do is determine the sign of the numerator to the left and right of t = -4 and t = -12.

The graph of the numerator (y = t² + 15t + 36) is a parabola that opens upward. It's pretty easy to see that for -12 < t < -4, y < 0, which means that f'(t) is negative on this interval. It's also easy to say that y > 0 for t < -12 or for t > - 4, meaning that f'(t) is positive on those two intervals.