Question Video: Finding the Expression of a Function given Its First Derivative and the Value of the Function at a Point | Nagwa Question Video: Finding the Expression of a Function given Its First Derivative and the Value of the Function at a Point | Nagwa

Question Video: Finding the Expression of a Function given Its First Derivative and the Value of the Function at a Point Mathematics • Second Year of Secondary School

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

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Video Transcript

Determine the function 𝑓 if the derivative 𝑓 dash of π‘₯ is equal to negative three π‘₯ plus one over the square root of π‘₯ and 𝑓 of one is equal to four.

To begin, let us rewrite 𝑓 dash of π‘₯ in a form that is easier to work with. Firstly, we can take out a factor of one over the square root of π‘₯. Next, we can recall that the square root of π‘₯ can be written as π‘₯ to the power of a half. We can take this one step further by recalling that π‘₯ to the power of a negative exponent, negative π‘Ž, is equal to one over π‘₯ to the π‘Ž.

We can therefore rewrite our term one over the square root of π‘₯ as π‘₯ to the power of negative a half. Let’s rewrite this in our equation and multiply out the terms. Looking at our first term, we can see that we have an π‘₯ to the power of one multiplied by an π‘₯ to the power of negative a half.

We can then recall that when multiplying exponents of the same variable π‘₯, we can simply add, or in this case subtract, the powers. Our term then becomes negative three π‘₯ to the power of one minus a half, in other words, negative three π‘₯ to the power of a half. Putting this term back into our equation gives us 𝑓 dash of π‘₯ in a more useful form.

We’re now ready to integrate 𝑓 dash of π‘₯ to find our function, 𝑓 of π‘₯. To start, we raise the power of our first term by one, from one half to three halves. We then divide by the new power. Next, we raise the power of our second term by one, from negative a half to a half. And we divide by the new power. Finally, we remember to add our constant of integration, 𝐢.

Let’s tidy things up. We can see that dividing by three over two is the same as multiplying by two over three. We can then cancel the three on the top and bottom half of this fraction, to give negative two π‘₯ to the power of three over two. We can also see that dividing by a half is the same as multiplying by two.

We have now found an equation for 𝑓 of π‘₯. However, since we have found the indefinite integral, this equation still contains the constant of integration, 𝐢. To find this constant, we can use the information given in the question, where 𝑓 of one is equal to four.

Let us rewrite 𝑓 of one using the equation that we have found. Here we can see our equation for 𝑓 of one. Since one raised to any power is also one, we can perform an easy simplification. We can now see that our negative two and positive two cancel out. We can therefore see that 𝐢 is equal to four.

Finally, we can rewrite 𝑓 of π‘₯ using the constant that we have just found. Our answer is therefore 𝑓 of π‘₯ equals negative two π‘₯ to the power of three over two add two π‘₯ to the power of a half add four. If you like, you could rewrite this term as two times the square root of π‘₯ to match the form given in the question.

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