Video: Finding the Value of a Function given the Expression of Its Slope by Using Indefinite Integration

Given that the slope at (π‘₯, 𝑦) is 3𝑒^(3π‘₯) and 𝑓(0) = βˆ’3, determine 𝑓(βˆ’3).

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Video Transcript

Given that the slope at the point π‘₯, 𝑦 is three 𝑒 to the power of three π‘₯ and 𝑓 evaluated at zero is negative three, determine 𝑓 evaluated at negative three.

In this question, we need to determine what 𝑓 evaluated at negative three is. And we’re told some information about our function 𝑓. We’re told that the slope at the point π‘₯, 𝑦 is given by three 𝑒 to the power of three π‘₯. And we’re also told that 𝑓 evaluated at zero is negative three. So we’re given two pieces of information about 𝑓 of π‘₯. First, we’re told the slope is three 𝑒 to the power of three π‘₯. And another way of saying this is 𝑓 prime of π‘₯ is equal to three 𝑒 to the power of three π‘₯. So to find 𝑓 of π‘₯, we need to find an antiderivative of 𝑓 prime of π‘₯. And we know how to do this by using integration.

We can find an antiderivative of 𝑓 prime of π‘₯ by integrating it with respect to π‘₯. We have 𝑓 of π‘₯ will be equal to the integral of three 𝑒 to the power of three π‘₯ with respect to π‘₯ up to a constant of integration. And to evaluate this integral, we need to recall the following. For any real constant π‘Ž, the integral of 𝑒 to the power of π‘Žπ‘₯ with respect to π‘₯ is equal to one over π‘Ž times 𝑒 to the power of π‘Žπ‘₯ plus a constant of integration 𝐢. We need to divide by the coefficient of π‘₯ in our exponent. And in our function, that’s equal to three. So we get three multiplied by one-third times 𝑒 to the power of three π‘₯. And remember, we need to add a constant of integration 𝐢.

Of course, we can simplify this. Three multiplied by one-third is just equal to one. So this simplifies to give us 𝑓 of π‘₯ is equal to 𝑒 to the power of three π‘₯ plus 𝐢. Now we want to find the value of 𝐢. And to do this, we need to use the fact that 𝑓 evaluated at zero is equal to negative three. So we substitue π‘₯ is equal to zero. We know 𝑓 of zero is negative three, and this is equal to 𝑒 to the power three times zero plus 𝐢. And now we just solved this equation for 𝐢. 𝑒 to the zeroth power is just equal to one. And then we rearrange to get 𝐢 is equal to negative four. So if 𝐢 is equal to negative four, we can substitute this into our expression for 𝑓 of π‘₯.

We have that 𝑓 of π‘₯ is equal to 𝑒 to the power of three π‘₯ minus four. The question wants us to find 𝑓 evaluated at negative three. So we substitute π‘₯ is equal to negative three into this expression. We get 𝑒 to the power of three times negative three minus four. And if we evaluate this expression and rearrange, we get our final answer of negative four plus one over 𝑒 to the ninth power.

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