We want to determine the points
where the given function has horizontal tangent lines.
Since horizontal lines have a slope of zero
and derivative function values
are equal to the slopes of tangent lines,
we will first determine where the given function
has horizontal tangent lines by finding the first derivative
and then setting the first derivative equal to zero,
and then solving for x.
The x, y's will give us the locations
at which the given function has horizontal tangent lines,
and then we'll have to determine
the corresponding function values
to determine the y coordinates for the points
where the function has horizontal tangent lines.
The function was given in the form here on the left.
Notice how I rewrote x cubed divided by three
as 1/3x cubed.
Let's use the form on the right to find the derivative.
f prime of x
is equal to the derivative of 1/3x cubed,
which is equal 1/3 times the derivative of x cubed,
which is 3x squared.
And we have minus the derivative of 3/2x squared,
which is equal to 3/2 times the derivative of x squared,
which is 2x, and then minus derivative of 10x,
which is 10 times the derivative of x,
which is one, giving it just minus 10,
minus the derivative of seven,
and the derivative of seven is zero.
Let's go ahead and simplify the derivative function.
Notice in the first product, the three is simplified to one.
In the second product, the two is simplified to one.
f prime of x, so the derivative function
is equal to x squared
minus 3x minus 10.
And again, now to find the x values
for which the given function has horizontal tangent lines,
we need to set f prime of x equal to zero and solve for x.
This gives us the equation x squared
minus 3x minus 10 equals zero.
We can solve this by factoring.
There will be two binomial factors on the left.
The factors of x squared are x and x.
The factors of -10 that add to -3 are -5 and +2.
The product on the left is equal to zero
when x minus five equals zero
or when x plus two equals zero,
which is when x equals five or x equals -2.
So we know the x coordinates for the points
where the given function has horizontal tangent lines.
One x coordinate is five
and the corresponding y coordinate is going to be f of five.
And the second point is -2, f of -2.
Now, so we're going to determine these two function values.
And just be careful, here we are subbing these x values
back to the original function in either form,
not the derivative function.
So f of five is equal to the cube of five
divided by three minus 3/2 times a square of five
minus 10 times five minus seven.
Let's also set up f of -2.
f of -2 is equal to the cube of -2
divided by three minus 3/2 times the square of -2
minus 10 times -2 minus seven.
And now, let's grow the calculator.
I've already set this up to save time.
Notice how the exact value of f of five is -317/6
or as a decimal, approximately -52.83.
f of -3 is exactly 13/3
or as a decimal, approximately 4.33.
Let's go ahead and record this.
Let's go ahead and use exact values for the y coordinates.
One point is 5,-317/6.
And the second point is -2,13/3.
These are the two points
where the given function has horizontal tangent lines.
And let's go ahead and verify this graphically.
The graph of the cubic function is in blue
Notice how there are two points
where the function has horizontal tangent lines.
On the left, we do have the point -2,13/3.
On the right, we do have the point 5,-317/6.
I hope you found this helpful.