Video: Finding the Expression of a Function given Its Second Derivative and the Value of the Function and Its Derivative at a Point Using Integration

Determine the function 𝑓 if 𝑓″(π‘₯) = βˆ’π‘₯⁴ + 1, 𝑓(1) = 5, and 𝑓′(1) = 3.

04:00

Video Transcript

Determine the function 𝑓 if 𝑓 double prime of π‘₯ is equal to negative π‘₯ 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 π‘₯ 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 π‘₯ is the same as saying the second derivative of 𝑓 with respect to π‘₯. And that’s the same as saying the derivative of d𝑓 dπ‘₯ with respect to π‘₯.

And the question tells us that this is equal to negative π‘₯ to the fourth power plus one. And we recall if we’re given a derivative function is equal to some function of π‘₯, we can solve this by using the opposite process to differentiation. We’re going to integrate both sides of this equation with respect to π‘₯. Remember, integrating is the opposite process to differentiation. So, integrating the derivative function of d𝑓 dπ‘₯ with respect to π‘₯ will just give us d𝑓 dπ‘₯. So, we have that d𝑓 dπ‘₯ is equal to the integral of negative π‘₯ to the fourth power plus one with respect to π‘₯.

We can rewrite d𝑓 dπ‘₯ as 𝑓 prime of π‘₯. 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 π‘₯ 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 π‘₯ to the zeroth power and then apply the power rule for integration. So, we now have that 𝑓 prime of π‘₯ is equal to negative π‘₯ to the fifth power over five plus π‘₯ 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 π‘₯ is equal to one into our equation for 𝑓 prime of π‘₯. 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 π‘₯ is equal to negative π‘₯ to the fifth power over five plus π‘₯ plus eleven-fifths.

But the question wants us to find the function 𝑓, not the function 𝑓 prime of π‘₯. So, to get from our derivative function 𝑓 prime of π‘₯ to 𝑓, we’re going to need to use the opposite process to differentiation. We’re going to need to integrate with respect to π‘₯. Using this and knowing that integration is the opposite process to differentiation, we have that 𝑓 of π‘₯ is equal to the integral of 𝑓 prime of π‘₯ with respect to π‘₯ which is also equal to the integral of negative π‘₯ to the fifth power over five plus π‘₯ plus 11 over five with respect to π‘₯. And again, we can evaluate this integral by using the power rule for integration.

Using the power rule for integration, we get negative π‘₯ to the sixth power over five times six plus π‘₯ squared over two plus eleven-fifths π‘₯ 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 π‘₯. 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 π‘₯ 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 π‘₯, we have shown that the function 𝑓 where 𝑓 double prime of π‘₯ is equal to negative π‘₯ 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 π‘₯ is equal to negative π‘₯ to the sixth power over 30 plus π‘₯ squared over two plus 11π‘₯ over five plus seven-thirds.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.