Show that the curve y=6x^3+5x-3 has no tangent line with slope 4.
The answer key says that
m=y'=18x^2+5, but x^2 is greater than or equal to 0 for all x, so m is greater than or equal to 5 for all x.
I don't understand that x^2 is greater than or equal to zero. Where did that come from? Why isn't x all real numbers?
y' = 18x^2 + 5
we want 18x^2 + 5 = 4
18x^2 = -1
x^2 = -1/18
but that is not possible since the square of anything has to be positive,
so there is value of x making this equations true
thus there is not slope of 4
thuse there is no such tangent
Well, let me break it down for you, but I'll do it with a touch of humor to make it more interesting.
So, we have the equation y = 6x^3 + 5x - 3, and we want to find the slope of the tangent line that has a slope of 4.
To find the slope, we need to take the derivative of y with respect to x, which gives us y' = 18x^2 + 5. This is the slope of the curve at any given point.
Now, let's take a closer look at this slope equation. We know that the square of any number is always greater than or equal to zero, right? Like, you can't have negative squared numbers, that would be some kind of weird math magic. So, x^2 is always greater than or equal to zero, no matter what value x is.
Since we want to find the points where the slope of the tangent line is 4, we set y' = 4 and solve for x. We have 18x^2 + 5 = 4. Now, we remember that 18x^2 is always greater than or equal to zero, so if we subtract 5 from both sides, we find that 18x^2 is greater than or equal to -1.
But wait! We said that x^2 is always greater than or equal to zero, right? So if 18x^2 is greater than or equal to -1, then x^2 must also be greater than or equal to -1/18. But since x^2 is always greater than or equal to zero, this means that x can't have any real values that satisfy the equation 18x^2 + 5 = 4.
Therefore, we conclude that the curve y = 6x^3 + 5x - 3 has no tangent line with a slope of 4. It's like trying to find a unicorn in a clown car, it just doesn't exist.
To show that the curve y = 6x^3 + 5x - 3 has no tangent line with slope 4, we need to find the derivative of the function and then analyze its values.
First, let's find the derivative of y with respect to x (dy/dx) to determine the slope of the tangent line at any given point:
dy/dx = 18x^2 + 5
To find the slope of the tangent line equal to 4, we set dy/dx equal to 4:
4 = 18x^2 + 5
Next, we solve this equation for x. Subtracting 5 from both sides gives:
18x^2 = -1
Dividing by 18:
x^2 = -1/18
From this equation, we can see that x^2 is less than zero, which means x does not have any real solutions. This implies that there is no x-value for which the curve has a tangent line with a slope of 4.
Hence, the curve y = 6x^3 + 5x - 3 has no tangent line with slope 4.
The statement that x^2 is greater than or equal to 0 for all x is incorrect. In fact, x^2 can take on negative values for certain x-values. In this case, the equation x^2 = -1/18 shows that x^2 is negative, which proves that there is no real solution for x.
To understand why x^2 is greater than or equal to zero, we can explore the concept of a square of a real number.
In general, when you raise any real number (positive or negative) to the power of 2 (x^2), the result is always non-negative or zero. This is because when you multiply a negative number by itself, the negative signs cancel out, resulting in a positive value. For positive numbers, multiplying them by themselves also yields a positive value. Therefore, the square of any real number is always greater than or equal to zero.
Now, let's apply this understanding to the given problem. We have a curve given by the equation y=6x^3+5x-3. To find the slope of the tangent line, we need to calculate the derivative of y with respect to x (y').
Taking the derivative of the equation, we get:
y' = 18x^2 + 5
To find out whether there exists a tangent line with a slope of 4, we need to determine the values of x for which the derivative y' is equal to 4.
Setting y' equal to 4, we have:
4 = 18x^2 + 5
Rearranging the equation, we get:
18x^2 = -1
Now, we can see that this equation doesn't have any real solutions because the left-hand side is always a non-negative value (since x^2 is always greater than or equal to zero), while the right-hand side is negative (-1). Therefore, there are no values of x that satisfy this equation, which means the curve has no tangent line with a slope of 4.