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Find Ine Intarval on whlch the graph &t f is cocave upward. Ine Interval on which the groph & {i5 corcave downward and ho coordinates ol the Iniecon points ...

Question

Find Ine Intarval on whlch the graph &t f is cocave upward. Ine Interval on which the groph & {i5 corcave downward and ho coordinates ol the Iniecon points .fl) x +122'_ Tr + [Answerthe inlervawhich ine graph of f is concave upNardIhe interval on which the grapn offis concave dowmtardand Ihe X, coordinates of the infectionpoints:_

Find Ine Intarval on whlch the graph &t f is cocave upward. Ine Interval on which the groph & {i5 corcave downward and ho coordinates ol the Iniecon points . fl) x +122'_ Tr + [ Answer the inlerva which ine graph of f is concave upNard Ihe interval on which the grapn offis concave dowmtard and Ihe X, coordinates of the infection points:_



Answers

Determine where the graph of the function is concave upward and where it is concave downward. $$ h(t)=\frac{t^{2}}{t-1} $$

We want to identify where function F is concrete. Up versus down F of X is equal to X squared over x minus one. This question is challenging understanding of the cavity. So let's first define the term conquering up versus down means that vegetable prime has signed positive versus negative at X. So we have to find a suitable prime F prime of x minus two X over x minus one square F. Double prime is two over x minus one cubed. We use where the numerator denominator acceptable prime. Zero to find the partition or critical points of vegetable prime that separates at double time intervals. For the denominator we have a partition a critical point at X equals one that we check the left and right of one for the sign of double prime left of one. Such as for example zero after the crime is negative To the right of one. Such as for x equals 10. Honourable prime is positive. Thus, given these signs of ethical prime on our two intervals, we have our solution. Our solution is Hafiz Khan came up on 1 to infinity, And if it's concave down on the interval between negative infinity and positive one.

We wanted to determine where the function is concave up or down. The function is F equals to t minus four to the power of one third. There are two steps executing this problem first, we're gonna find the second derivative F double prime and zeros in a sentence. These zeros innocent toast will separate F double prime and F into a series of intervals along the X axis. For stuff to We're going to evaluate the science of prime on each of these intervals wherever F double prime is positive, the function is concave up wherever F double prime is negative. Fx concave down. So first let's find the derivatives and the intervals in question. So F prime is just two thirds to t minus 40. 2 3rd. F double prime is negative. 8/9 to t minus four to negative five thirds. Dysfunction does not have a zero but it does have a break or innocent at T equals two. So we have an interval negative affinity to and to to infinity on these intervals, F double prime is for negativity to negative and from two to infinity positive. So we have solution Fs concave up from negative infinity to and concrete down from two to infinity.

We want to identify with function F. It's concrete. Up versus down F of X is equal to two x minus four. The one third to answer this question, we identify what can cavity means. So remember that concrete up versus down means that double prime is positive versus negative for X. F. Prime is two thirds over two x minus four to the two thirds after the prime is negative. 8/9. 2 x minus four. Of the five thirds we find critical or petition points inevitable prime. It has no critical points but from its denominator and has a partition and X equals two. That's we have to evaluate the sign of a double prime left and right of X equals studio identify at this concrete up versus down left of X equals two. X equals 10. After the primer is positive to the rate of X equals two. Such as an excuse to an actual crime is negative. Thus we have our solution F as khan came up from negative three or two where after the parliament is negative, rather positive And f has gone came down on the interval from 2 to infinity.

All right. The theme in this problem is that uh we want the secondary to find intervals of common. Cave up khan came down. So we need to find the second derivative of this problem. So, uh just double check in, typed it correctly because the first year of debate, I just need in order to get the second derivative. What we learn in this section is the contributed and the integral will cancel each other out. You just have to plug in X. And for that T. And then what we need next is the derivative of that. And that's going to be your chain rule. Because there's an inner function. Don't write that down because the derivative of the outside the outer function E. Part is itself. But then you have to multiply by the derivative of the expo, which would be -2 X. Um So now what we need to do for intervals of con cavity is to compare this to zero. Um Now what some students will forget is that e to any power is always going to be a positive. That's the property of exponents. Um so the only time that this change is kind of cavity where the secondary of the equal zero is if X. Y two equals zero. Because anything times negative two times zero gives me zero. So I know some people won't let you just um make a number line and talk about, well, if X were and negative number again, this is always positive and then you have a negative times a negative, which means that you'll get a positive answer. Uh For the second derivative right have double prime. Um Usually feel a greater than zero instead of a positive. But what that means is that you have a concave up Graph from Negative Infinity to zero. I. Circle that. So what if we have a positive version of ex again this is always positive and then a negative times of possible. Give me a negative, so therefore the second derivative will be less than zero. And when that happens we are con caved down On that interval from 0 to infinity. Now there are two answers circled in green, that's what we want.


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