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.
