Each limit represents the derivative of some function f at some number a. Indicate an f and a in each instance.limt→1t4−t−2t−1

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Aretha Frazier · Accepted Answer

Step 1Consider the limitlimt→1t4−t−2t−1We want to represent it as a derivative of some function f at some number a, ln⁡ light of definition 5 we want to write our limit in the formlimt→af(t)−f(a)t−aSince t−1 appears in the denominator, and t→1, we are motivated to take a=1Looking at the numerator of the limit in question we want f(t)=t4+t. Let us see if this is the right choice. By definition 5 we havef′(1)=limt→1f(t)−f(1)t−1=limt→1t4+t−(14+1)t−1=limt→1t4+t−2t−1Which is what we wanted to establish.Ans: f(t)=t4+t, a=1

Hiroko Cabezas · Accepted Answer

Step 1Definitionf′(a)=limx→af(x)−f(a)x−aStep 2limt→1t4+t−2t−1limt→1(t4+t)−(14+1)t−1limt→1f(t)−f(1)t−1=f′(1)Where f(t)=t4+tNote that a=1

Andre BalkonE · Accepted Answer

To find the function and number associated with the given limit limt→1t4−t−2t−1, we can rewrite it using the derivative notation. Let&#039;s start by simplifying the expression:limt→1t4−t−2t−1Factoring the numerator as (t−2)(t+1)(t2+1), we can cancel out the common factor of (t−1) in both the numerator and denominator:limt→1(t−2)(t+1)(t2+1)t−1Now, we can see that the limit resembles the derivative of a function. To find the function f and number a associated with this limit, we rewrite it as:limt→1(t−2)(t+1)(t2+1)t−1=limt→1(t+1)(t2+1)By observing the expression inside the limit, we can identify the function f(t)=t2+1 and the number a=1. Therefore, the given limit represents the derivative of the function f(t)=t2+1 at t=1.

Jazz Frenia · Accepted Answer

Answer:f(t)=t4+t, a=1Explanation:limt→1t4−t−2t−1To simplify this expression, we can factor the numerator using the difference of cubes formula:t4−t−2=(t2+1)(t2−2)Now we can rewrite the limit as:limt→1(t2+1)(t2−2)t−1Since the denominator is t−1, we can cancel it out with the factor (t2−2) in the numerator:limt→1(t2+1)(t+1)Next, we can evaluate the limit as t approaches 1. Plugging in t=1 into the expression, we get:(12+1)(1+1)=2(2)=4Therefore, the function f(t) that represents the given limit is f(t)=t4+t, and the number a is a=1.

fudzisako · Accepted Answer

Given limit: limt→1t4−t−2t−1To simplify the expression, we can factorize the numerator:(t−2)(t+1)(t2+2t+1)t−1Now, cancel out the common factor of (t−1) from the numerator and the denominator:(t−2)(t+1)(t+1)t−1Simplifying further, we get:(t−2)(t+1)(t+1)Now, let&#039;s find a function f(t) whose derivative matches this expression. We can take:f(t)=(t−2)33To find the value of a, we substitute t=1 into the function:f(1)=(1−2)33=−13Therefore, the function f(t) and the value of a for the given limit are:f(t)=(t−2)33 and a=1So, the solution can be written as:limt→1t4−t−2t−1=limt→1(t−2)(t+1)(t+1)=f′(1), where f(t)=(t−2)33 and f′(t) represents the derivative of f(t).

Each limit represents the derivative of some function f at

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