Let f(x) be a polynomial function such that f(-5)=-2, f'(-5)=0, and f''(-5)=6. Classify the point (-5,-2).
a) intercept b) inflection point c) local minimum d) local maximum e) none of the above
First derivative zero means a critical point; second derivative positive means local minimum.
Right, I understand that. But as far as classifying the point (-5,-2)?
See, I thought that(-5,-2) was an intercept point. If x is -5, and f(-5) gives you -2, wouldn't that make -2 the u intercept? And thereby making (-5,-2) intercept points?
*y intercept I meant
Well the slope at that point is 0. So this tells that the tangent line to the curve that hits this point is a horizontal line. The y-intercept is when x=0 tho
Well, the value of the function at x=-5 is y=-2. The other info means at that x value, the function has a local minimum.
Oh, yes, that's right. Y int at x=0. Sorry bout that.
hmm, but if the first derivative is a horizontal line then the second derivative wouldn't have a value of 6. This is a tough question
Yeah, I got it wrong on my final exam. I'm a bit bummed about it.
The second derivative tells how the slope is changing. If the slope is increasing (as it will when the second derivative is positive) there will be a minimum value at the point where the first derivative is zero.
Well, I think Animal is saying it is Local Min. Right, Animal? I don't have the correct answer, just my wrong answer.
Yeah. Local minimum is correct.
Intercept. It's not that one.
So because -5 is a local minimum, that classifies the entire point (-5,-2) as a local minimum?
The local minimum happens at x=-5; the value of the function at that point is y=-2. So, you might say the function has a local minimum at x=-5, or the local minimum value of the function is -2 in the neighborhood of x=-5.
Ok, that makes sense. Thanks for clearing it up.
No sweat. Do math every day.
Join our real-time social learning platform and learn together with your friends!