# Question Video: Finding the Local Maximum and Minimum Values of a Function given the Slope of Its Tangent and a Point on the Curve Involving Using Integration Mathematics

Find the local maximum and minimum values of the curve that passes through the point (−1, 7) where the gradient of the tangent is 6(𝑥² + 4𝑥 + 3).

04:24

### Video Transcript

Find the local maximum and minimum values of the curve that passes through the point negative one, seven where the gradient of the tangent is six times 𝑥 squared plus four 𝑥 plus three.

In this question, we need to find the local maximum and minimum values of a curve, but we’re not given the equation for our curve directly. Instead, we’re only given information about our curve. We’re told that our curve passes through the point negative one, seven. And we’re also given the gradient of the tangent line at the point 𝑥, 𝑦. It’s given by six times 𝑥 squared plus four 𝑥 plus three. Of course, to find the local maximum and minimum values of a function, we’re going to need to know its output at these values. So we’re going to want to find an expression for our function. And to do this, we’re going to need to use the two pieces of information we’re given about our curve.

First, we’re given an expression which represents the gradient of the tangent lines to our curve at a value of 𝑥. And we recall this is exactly the same as saying the rate of change of 𝑦 with respect to 𝑥 is equal to this expression. In other words, we have d𝑦 by d𝑥 is equal to six times 𝑥 squared plus four 𝑥 plus three. But this is not the only piece of information we’re given. We’re also told that our curve passes through the point negative one, seven. So when 𝑥 is equal to negative one, we must have that 𝑦 is equal to seven. We’re given an expression for the gradient of our curve.

So differentiating our curve with respect to 𝑥, we get the following expression. This means we want to find an antiderivative of this expression. And we know how to do this by using integration. We must have that 𝑦 is equal to the integral of six times 𝑥 squared plus four __ plus three with respect to 𝑥 up to a constant of integration because this will give us the most general antiderivative of this expression. There’s a few different ways of solving this. We’re going to distribute six over our parentheses. Doing this, we get the integral of six 𝑥 squared plus 24𝑥 plus 18 with respect to 𝑥. And now this is just the integral of a quadratic. We can do this term by term by using the power rule for integration.

We recall for any real constants 𝑎 and 𝑛, where 𝑛 is not equal to negative one, the integral of 𝑎𝑥 to the 𝑛th power with respect to 𝑥 is equal to 𝑎 times 𝑥 to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝐶. We add one to our exponent of 𝑥 and then divide by this new exponent. Applying this to six 𝑥 squared, we add one to the exponent of two giving us a new exponent of three and then divide by the exponent of three. Doing this and simplifying, we get two 𝑥 cubed. We’ll do the same for 24𝑥, which we can write as 24𝑥 to the first power. We add one to our exponent of one and then divide by this new exponent. This gives us 12𝑥 squared. Finally, we could do the same for 18 by writing it as 18 times 𝑥 to the zeroth power. However, we can just write the antiderivative of this as 18𝑥. And don’t forget, we need to add our constant of integration 𝐶.

So now we found an expression for the curve given to us in the question up to a constant of integration 𝐶. We now want to find the value of our constant 𝐶. And to do this, we’re going to need to use the fact that our curve passes through the point negative one, seven. If our curve passes through this point, we can substitute these into the equation for our curve. And then the equation has to be true for our curve to pass through this point. Substituting 𝑥 is equal to negative one and 𝑦 is equal to seven into the equation for our curve, we get seven is equal to two times negative one cubed plus 12 times negative one squared plus 18 times negative one plus 𝐶.

Now by simplifying this expression and solving for 𝐶, we get that 𝐶 is equal to 15. And if 𝐶 is equal to 15, we can substitute this into the expression for our curve. So by clearing some space and using 𝐶 is equal to 15, we have the following expression for our curve. But remember, we need to find the local maximum and minimum values of this curve. And to find this, we need to recall the following piece of information. In polynomials, our local extrema will occur at the turning points, in other words, when the gradient is equal to zero. And we’re given an expression for the gradient.

So to find our local extrema, we need to solve six times 𝑥 squared plus four 𝑥 plus three is equal to zero. And to do this, all we need to do is factor our quadratic. We see this factors to give us 𝑥 plus one times 𝑥 plus three. Solving this is equal to zero, we get 𝑥 is negative one or 𝑥 is negative three. To find the values of these, we need to substitute these into the equation for our curve. Substituting negative one into our curve, we get 𝑦 is equal to seven. Similarly, when 𝑥 is equal to negative three, we see that 𝑦 is equal to 15. We now need to determine which of these is a maximum and which is a minimum. To do this, we notice our curve is a cubic with a positive leading coefficient. It will look something like this. In fact, we know it has two turning points, and we know the coordinates of these two turning points. So we’ve shown the local maximum value is 15 and the local minimum value is seven.

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