Find the largest interval on which f is increasing

[msrultons]

Homework Statement

find the largest interval on which f is increasing

Relevant Equations

Fundamental Theorem of Calculus

Attached here is the full problem I am doing.

I went through the problem and got my final answer which I thought was correct. Here is my work. They tell me I am wrong. Not sure where is the mistake.

 

[msrultons]

Your hand-written answer looks good to me. Did you type in the answer in the first photo? That is wrong.

I tried both -1,1 and then was trying out other answers to see what the program was selecting.

 

[FactChecker]

Should you include the endpoints to get the "largest" interval?

 

[Mark44]

Maybe the problem is distinguishing between increasing and strictly increasing. For f increasing (i.e., $f' \ge 0$), the interval is [-1, 1]. For strictly increasing ($f' > 0$), the interval is (-1, 1).

 

[fresh_42]

You should consider typing your solutions instead of uploading pictures.
Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

 

[Office_Shredder]

Maybe the problem is distinguishing between increasing and strictly increasing. For f increasing (i.e., $f' \ge 0$), the interval is [-1, 1]. For strictly increasing ($f' > 0$), the interval is (-1, 1).

f can be strictly increasing on an interval where its derivative is zero. For example $x^2$ on $[0,\infty)$, the only time it's zero is at zero so you can include it in the interval and it's still strictly increasing. It's actually only degenerate examples where you can't include it.

 

[Mark44]

I was probably conflating the concept of an increasing/strictly increasing function vs. the derivatives here. It's very possible that the software was looking for [-1, 1] instead of (-1, 1).

 

[TeethWhitener]

If $x\in\mathbb{R}$, you have to consider negative values of $x$, which has an important effect on the sign of the integral.

 

[Mark44]

If $x\in\mathbb{R}$, you have to consider negative values of $x$, which has an important effect on the sign of the integral.

I don't see that x being negative is relevant here. It seems clear to me that to find the interval on which f is increasing, you want the derivative of the given integral, which as the OP wrote, is $f'(x) = (1 - x^2)e^{x^4}$. And we want to know the interval for which this derivative is >= 0.

 

[TeethWhitener]

$f(x)$ in OP is dependent on the limits of integration. $\int_a^b {g(x){}dx}=-\int_b^a{g(x){}dx}$

Edit: a little more on the nose: $f’(x)\neq(1-x^2)e^{x^4}$

 

[Mark44]

$f(x)$ in OP is dependent on the limits of integration. $\int_a^b {g(x){}dx}=-\int_b^a{g(x){}dx}$

That's a good point, but for an integral of the form $\int_a^b f(x) dx$, a is usually less than b. If they wrote the integral in post #1 where x < 0, that would be especially sneaky of the writers for a problem that apparently checks whether the solver understands the Fundamental Thm. of Calculus.

 

[TeethWhitener]

That's a good point, but for an integral of the form $\int_a^b f(x) dx$, a is usually less than b. If they wrote the integral in post #1 where x < 0, that would be especially sneaky of the writers for a problem that apparently checks whether the solver understands the Fundamental Thm. of Calculus.

The domain of $f(x)$ is unspecified in the problem. It could be $[\frac{1}{2},\frac{3}{2}]$ for all I know, so I can’t answer the problem with certainty. That’s why I gave the condition of $x\in\mathbb{R}$ in post 8. I suppose it’s sneaky, but I don’t think it’s out of line in a first year calculus course to consider what happens when the upper limit of integration is less than the lower limit.

 

[Office_Shredder]

If x is less than zero it makes computing the integral slightly trickier, but it doesn't affect the value of the derivative at all. The fundamental theorem of calculus is not dependent on your choice of the lower bound of the integral