Video Transcript
Determine the function 𝑓 if 𝑓 double prime of 𝑥 is equal to negative 𝑥 to the fourth power plus one, 𝑓 evaluated at one is equal to five, and 𝑓 prime evaluated at one is equal to three.
The question is asking us to find the function 𝑓, given that the second derivative of 𝑓 is equal to negative 𝑥 to the fourth power plus one, 𝑓 evaluated at one is five, and the first derivative of 𝑓 evaluated at one is equal to three. We recall that 𝑓 double prime of 𝑥 is the same as saying the second derivative of 𝑓 with respect to 𝑥. And that’s the same as saying the derivative of d𝑓 d𝑥 with respect to 𝑥.
And the question tells us that this is equal to negative 𝑥 to the fourth power plus one. And we recall if we’re given a derivative function is equal to some function of 𝑥, we can solve this by using the opposite process to differentiation. We’re going to integrate both sides of this equation with respect to 𝑥. Remember, integrating is the opposite process to differentiation. So, integrating the derivative function of d𝑓 d𝑥 with respect to 𝑥 will just give us d𝑓 d𝑥. So, we have that d𝑓 d𝑥 is equal to the integral of negative 𝑥 to the fourth power plus one with respect to 𝑥.
We can rewrite d𝑓 d𝑥 as 𝑓 prime of 𝑥. We can then integrate this by using the power rule for integration. We add one to the exponent and then divide by this new exponent. To integrate negative 𝑥 to the fourth power, we add one to four to get five and then divide by this new value of five. And to integrate one, we rewrite it as 𝑥 to the zeroth power and then apply the power rule for integration. So, we now have that 𝑓 prime of 𝑥 is equal to negative 𝑥 to the fifth power over five plus 𝑥 plus the constant of integration 𝑐 one.
We recall the question tells us that 𝑓 prime evaluated at one is equal to three. So, we can use this to find the value of 𝑐 one. We substitute 𝑥 is equal to one into our equation for 𝑓 prime of 𝑥. This gives us three is equal to negative one to the fifth power over five plus one plus 𝑐 one. Rearranging and simplifying this, we see that 𝑐 one is equal to eleven-fifths. So, we’ve shown that 𝑓 prime of 𝑥 is equal to negative 𝑥 to the fifth power over five plus 𝑥 plus eleven-fifths.
But the question wants us to find the function 𝑓, not the function 𝑓 prime of 𝑥. So, to get from our derivative function 𝑓 prime of 𝑥 to 𝑓, we’re going to need to use the opposite process to differentiation. We’re going to need to integrate with respect to 𝑥. Using this and knowing that integration is the opposite process to differentiation, we have that 𝑓 of 𝑥 is equal to the integral of 𝑓 prime of 𝑥 with respect to 𝑥 which is also equal to the integral of negative 𝑥 to the fifth power over five plus 𝑥 plus 11 over five with respect to 𝑥. And again, we can evaluate this integral by using the power rule for integration.
Using the power rule for integration, we get negative 𝑥 to the sixth power over five times six plus 𝑥 squared over two plus eleven-fifths 𝑥 plus the constant of integration 𝑐 two. We can simplify five times six to just give us 30. So, we’ve now found an expression for our function 𝑓 of 𝑥. However, it still has the unknown constant 𝑐 two. And we recall the question told us that 𝑓 evaluated at one was equal to five. Using this and substituting 𝑥 is equal to one gives us that five is equal to negative one times one to the sixth power over 30 plus one squared over two plus eleven-fifths times one plus 𝑐 two. Rearranging, simplifying, and evaluating this expression, we can see that 𝑐 two is equal to seven over three.
If we then substitute this value of 𝑐 two back into our expression for 𝑓 of 𝑥, we have shown that the function 𝑓 where 𝑓 double prime of 𝑥 is equal to negative 𝑥 to the fourth power plus one and 𝑓 evaluated at one is equal to five and 𝑓 prime of one is equal to three. Is given by 𝑓 of 𝑥 is equal to negative 𝑥 to the sixth power over 30 plus 𝑥 squared over two plus 11𝑥 over five plus seven-thirds.
