# Question Video: Finding the Local Minimum Value of a Function given Its Local Maximum Value and the Expression of Its Derivative Involving Using Integration Mathematics

Find the local minimum value of a curve given that its gradient is d𝑦/d𝑥 = 𝑥² + 3𝑥 − 18 and the local maximum value is 21.

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### Video Transcript

Find the local minimum value of a curve given that its gradient is d𝑦 by d𝑥 is equal to 𝑥 squared plus three 𝑥 minus 18 and the local maximum value is 21.

In this question, we’re tasked with finding the local minimum value of a curve. We’re given an equation for its slope: d𝑦 by d𝑥 is 𝑥 squared plus three 𝑥 minus 18. And we’re told the local maximum value is 21. To answer this question, we start by recalling that the local extrema of a function will always occur at the critical points of a function. And those are the points where the derivative of the function is zero or the derivative does not exist. And we’re given the gradient of this function. It’s 𝑥 squared plus three 𝑥 minus 18. This is a quadratic polynomial. And quadratic functions are defined for all real values of 𝑥. So, the gradient of this function is defined for all values of 𝑥. The only critical points will occur when the gradient is zero.

So, let’s determine the 𝑥-values of the critical points of this function. We need to solve 𝑥 squared plus three 𝑥 minus 18 is equal to zero. And we can do this by factoring. We notice that six multiplied by negative three is negative 18 and six plus negative three is equal to three. Therefore, we can factor this quadratic to get 𝑥 plus six multiplied by 𝑥 minus three is equal to zero. And for a product to be equal to zero, one of the two factors must be equal to zero. In other words, either 𝑥 is negative six or 𝑥 is equal to three. These are the two critical points of our curve.

However, this is not enough to determine the local minimum value of this curve. All we know is it either occurs when 𝑥 is negative six or occurs when 𝑥 is equal to three. To determine this value, we’re going to need to find an equation of our curve. And we can do this by using the fact we’re given an equation for the gradient of this curve. Therefore, 𝑦 will be an antiderivative of its gradient. 𝑦 is equal to the indefinite integral of d𝑦 by d𝑥 with respect to 𝑥. And this is, of course, up to a constant of integration. Let’s substitute the expression we’re given for d𝑦 by d𝑥 into our integral. We get that 𝑦 is equal to the indefinite integral of 𝑥 squared plus three 𝑥 minus 18 with respect to 𝑥.

Since this is the integral of a polynomial, we can do this term by term by using the power rule for integration, which we recall tells us for any real constants 𝑎 and 𝑛, where 𝑛 is not 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 can apply this term by term and then add a constant of integration at the end of this expression. We get 𝑥 cubed over three plus three 𝑥 squared over two minus 18𝑥 plus 𝐶. This is the equation of our curve for some value of 𝐶.

And because we don’t know the value of 𝐶, we can’t determine the local minimum value of this curve. It will change depending on the value of 𝐶. And we can’t just substitute 𝑥 is negative six or 𝑥 is three into this curve because we don’t know the corresponding 𝑦-coordinate. So, we need to determine the coordinates of a point which lies on the curve. This will allow us to find the value of 𝐶. To do this, we need to use the fact in the question that we’re told the local maximum value of this curve is 21. And since this is a local extrema, we know this will occur at one of the critical points. It either occurs when 𝑥 is negative six or when 𝑥 is equal to three. Therefore, we need to determine which of these two critical points is the local maximum.

There’s several different ways of doing this. For example, we could use the first derivative test. However, we’re going to use the fact that we know 𝑦 is a cubic polynomial. And we can see that this has positive leading coefficient. And finally, we know that the curve has two critical points. In other words, we can sketch the shape of the function. It’s a positive leading coefficient cubic polynomial with two turning points. And the first turning point has 𝑥-coordinate negative six, and the second turning point has 𝑥-coordinate three. Therefore, when 𝑥 is negative six, our curve has a local maximum, and when 𝑥 is three, our curve has a local minimum. This then allows us to find the coordinates of a point which lies on the curve. The local maximum value of this curve is 21. So, when 𝑥 is negative six, 𝑦 is 21.

Therefore, we can substitute 𝑥 is negative six and 𝑦 is 21 into the equation of our curve to find the value of 𝐶. This gives us that 21 is equal to negative six cubed over three plus three times negative six squared over two minus 18 multiplied by negative six plus 𝐶. We can then evaluate and rearrange to find the value of 𝐶. We get that 𝐶 is equal to negative 69. We can then substitute this value of 𝐶 into the equation of our curve. This gives us that the equation of our curve is 𝑦 is equal to 𝑥 cubed over three plus three 𝑥 squared over two minus 18𝑥 minus 69.

And now we can find the value of the local minimum of this curve. We know that this occurs at the critical point 𝑥 is equal to three. So, we substitute 𝑥 is equal to three into the equation of our curve. We get 𝑦 is equal to three cubed over three plus three times three squared over two minus 18 times three minus 69. We can then evaluate this expression to find the 𝑦-coordinate of this point is negative 100.5, which is our final answer.

Therefore, we were able to show if the gradient of a curve is given by d𝑦 by d𝑥 is equal to 𝑥 squared plus three 𝑥 minus 18 and the local maximum value of this curve is 21, then the local minimum value is negative 100.5.