Video Transcript
The slope of the tangent to the curve 𝑦
is equal to 𝑓 of 𝑥 at the point 𝑥, 𝑦 is given by three 𝑥𝑒 to the two 𝑥 divided by two
𝑥 plus one squared. Determine the function 𝑓 of 𝑥 if the
point one, five 𝑒 squared lies on the curve.
The question tells us the slope to the
curve 𝑦 is equal to 𝑓 of 𝑥 at the point 𝑥, 𝑦 is given by three 𝑥𝑒 to the power of two
𝑥 all divided by two 𝑥 plus one squared. In other words, we’re told 𝑓 prime of 𝑥
is equal to three 𝑥𝑒 to the two 𝑥 divided by two 𝑥 plus one squared. The question wants us to determine the
function 𝑓 of 𝑥 given that the point one, five 𝑒 squared lies on our curve 𝑦 is equal to
𝑓 of 𝑥.
To do this, we’re going to notice that 𝑓
of 𝑥 is an antiderivative of its slope function 𝑓 prime of 𝑥. In other words, our function 𝑓 of 𝑥 is
equal to the integral of 𝑓 prime of 𝑥 with respect to 𝑥 up to our constant of integration
𝑐. So, we need to integrate three 𝑥𝑒 to
the power of two 𝑥 divided by two 𝑥 plus one squared with respect to 𝑥. We notice that our integrand is not in a
standard form which we can integrate. So, we’re going to need to perform some
kind of manipulation.
We could try using the substitution 𝑢 is
equal to two 𝑥 plus one. However, this does not make current to go
any easier. So, we’re going to try using integration
by parts. We recall integration by parts tells us
the integral of 𝑢𝑣 prime with respect to 𝑥 is equal to 𝑢𝑣 minus the integral of 𝑢
prime 𝑣 with respect to 𝑥. It might be tempting at this point to try
and set our function 𝑢 to be some kind of algebraic function. However, we’re going to set our function
𝑢 to be the numerator of our fraction, three 𝑥 multiplied by 𝑒 to the power of two
𝑥.
We’ll also write 𝑣 prime to be one
divided by the denominator of our fraction. We’ll write this as two 𝑥 plus one all
raised to the power of negative two. To calculate 𝑢 prime, we’re going to use
the product rule. First, we differentiate three 𝑥 to get
three. Then, we multiply this by 𝑒 to the power
of two 𝑥. Then, we add on the derivative of 𝑒 to
the power of two 𝑥. That’s two 𝑒 to of the power of two
𝑥. And we multiply this by three 𝑥.
Now, we can notice that both terms in our
expression for 𝑢 prime share a factor of three 𝑒 to the power of two 𝑥. So, we can factor this out, giving us
three 𝑒 to the two 𝑥 multiplied by one plus two 𝑥. To find our antiderivative of the
function 𝑣 prime, we’re going to use integration by substitution. We’ll use the substitution 𝑤 is equal to
two 𝑥 plus one. Differentiating both sides of this
equation with respect to 𝑥 gives us that the derivative of 𝑤 with respect to 𝑥 is equal
to two.
And although d𝑤 by d𝑥 is not a
fraction, when using integration by substitution, it behaves a little bit like one. This gives us the equivalent statement
that half d𝑤 is equal to d𝑥. So, by using the substitution 𝑤 is equal
to two 𝑥 plus one, we have that 𝑣 is equal to the integral of 𝑤 raised to the power of
negative two multiplied by a half with respect to 𝑤. We can integrate this by adding one to
our exponent and then dividing by the new exponent. This gives us that 𝑣 is equal to
negative 𝑤 to the power of negative one divided by two.
Finally, by substituting 𝑤 is equal to
two 𝑥 plus one and bringing 𝑤 down into our denominator, we have that 𝑣 is equal to
negative one divided by two multiplied by two 𝑥 plus one. So, by using integration by parts, we’ve
shown that our function 𝑓 of 𝑥 is equal to 𝑢 multiplied by 𝑣, which is three 𝑥𝑒 to the
power of two 𝑥 multiplied by negative one divided by two multiplied by two 𝑥 plus one. And then, we subtract the integral of 𝑢
prime multiplied by 𝑣 with respect to 𝑥, which is the integral of negative one divided by
two multiplied by two 𝑥 plus one all multiplied by three 𝑒 to the power of two 𝑥
multiplied by one plus two 𝑥 with respect to 𝑥.
At first, this might not seem any
simpler. However, we can notice in our integrand
that our denominator and our numerator share a factor of one plus two 𝑥. So, we can simplify this to be negative
three 𝑥𝑒 to the power of two 𝑥 divided by two multiplied by two 𝑥 plus one minus the
integral of negative three 𝑒 to the two 𝑥 over two with respect to 𝑥. We can now evaluate this integral by
recalling that for contestants 𝑎 and 𝑛, the integral of 𝑎𝑒 to the 𝑛𝑥 with respect to
𝑥 is equal to 𝑎𝑒 to the 𝑛𝑥 divided by 𝑛 plus our constant of integration 𝑐.
So, evaluating the negative of this
integral gives us three 𝑒 to the power of two 𝑥 divided by four plus our constant of
integration 𝑐. We can simplify this further by taking
out a shared factor of three-quarters 𝑒 to the power of two 𝑥 from our first and second
term. This gives us three-quarters 𝑒 to the
power of two 𝑥 multiplied by negative two 𝑥 over two 𝑥 plus one plus one plus our
constant of integration 𝑐.
We can then add the terms inside our
parentheses by noticing that one is equal to two 𝑥 plus one over two 𝑥 plus one. Then, we see that the negative two 𝑥 and
the two 𝑥 in our numerators cancel. So, we’ve shown that our function 𝑓 of
𝑥 is equal to three-quarters 𝑒 to the power of two 𝑥 multiplied by one over two 𝑥 plus
one plus our constant of integration 𝑐.
Finally, we recall that the question
tells us that the point one, five 𝑒 squared lies on the curve 𝑦 is equal to 𝑓 of 𝑥. If the point one, five 𝑒 squared lies on
our curve 𝑦 is equal to 𝑓 of 𝑥, then when 𝑥 is equal to one, 𝑦 must be equal to five 𝑒
squared. So, we substitute 𝑥 is equal to one and
𝑦 is equal to five 𝑒 squared into our equation. This gives us that five 𝑒 squared is
equal to three-quarters 𝑒 to the power of two times one all multiplied by one over two
times one plus one plus our constant of integration 𝑐.
The right-hand side of our equation
simplifies to give us three-quarters 𝑒 squared multiplied by a third plus 𝑐. We can then cancel the shared factors of
three in our numerator and our denominator, giving us that five 𝑒 squared is equal to 𝑒
squared over four plus 𝑐. Finally, we’ll subtract 𝑒 squared over
four from both sides of this equation to give us that 𝑐 is equal to 19𝑒 squared over
four.
Therefore, after a bit of rearranging,
we’ve shown that if the slope of the tangent to the curve 𝑦 is equal to 𝑓 of 𝑥 at the
point 𝑥, 𝑦 is given by three 𝑥𝑒 to the two 𝑥 over two 𝑥 plus one squared and our
curve passes through the point one, five 𝑒 squared, then, our function 𝑓 of 𝑥 is equal to
three 𝑒 to the power of two 𝑥 over four multiplied by two 𝑥 plus one plus 19𝑒 squared
over four.
