Describing Points - CLEP Calculus
Card 0 of 4
Suppose a point on the curve given above has the property that 
.
Based solely on the graph above, which of the following is most likely the 
value of the point in question?
Suppose a point on the curve given above has the property that
Based solely on the graph above, which of the following is most likely the
If
then the graph must be concave up at the point. Based on the picture, we know that the curve is concave up on 
at best. The only value that falls on this interval is 
, which is 
. Since 
, this definitely falls on the interval given and we can be sure it is concave up based on the picture.
If
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What is the critical point for 
?
What is the critical point for
To find the critical point, you must find the derivative first. To do that, multiply the exponent by the coefficient in front of the 
and then subtract the exponent by 
. Therefore, the derivative is: 
. Then, to find the critical point, set the derivative equal to 
.
.
To find the critical point, you must find the derivative first. To do that, multiply the exponent by the coefficient in front of the
Compare your answer with the correct one above
Suppose a point on the curve given above has the property that .
Based solely on the graph above, which of the following is most likely the value of the point in question?
Suppose a point on the curve given above has the property that
Based solely on the graph above, which of the following is most likely the
If then the graph must be concave up at the point. Based on the picture, we know that the curve is concave up on at best. The only value that falls on this interval is , which is . Since , this definitely falls on the interval given and we can be sure it is concave up based on the picture.
If
Compare your answer with the correct one above
What is the critical point for ?
What is the critical point for
To find the critical point, you must find the derivative first. To do that, multiply the exponent by the coefficient in front of the and then subtract the exponent by . Therefore, the derivative is: . Then, to find the critical point, set the derivative equal to .
.
To find the critical point, you must find the derivative first. To do that, multiply the exponent by the coefficient in front of the
Compare your answer with the correct one above