5. Let f be the function given by f(x) = x3- 7x + 6.
a. Find the zeros of f
b. Write an equation of the line tangent to the graph of f at x = -1
c. Find the x coordinate of the point where the tangent line is parallel to the secant line on the interval [1, 3].
Well, first off, if x=1 we have a zero.
So divide by (x-1) and get
x^2+x-6
factor that
(x-2)(x+3)
so the zeros are at x = 1, 2, -3
At x - -1
y = -1 + 7 + 6 = 12
y' = 3 x^2 - 7 which is -4 at x = -1
so this line goes through (-1,12) and has slope m = -4
You can do the rest
at x = 1, y = 0
at x = 3, y = 27-21+6 = 12
line slope m = 12/2 = 6
where is slope of tangent line = 6?
3x^2 -7 = 6
3 x^2 = 13
x^2 = 13/3
x = sqrt (13/3)
a. Well, to find the zeros of the function f(x) = x^3 - 7x + 6, we need to solve for x when f(x) = 0. This means we set the equation x^3 - 7x + 6 = 0. But let me tell you, this equation is being quite stubborn, it doesn't want to give up its zeros easily!
b. Now, to find the equation of the tangent line at x = -1, we need to find the slope of the tangent line first. Let me tell you a secret, finding slopes can sometimes be as slippery as a banana peel! But once we find the slope, we can use the point-slope form to get the equation of the line.
c. Ah, the parallelism between lines! It's like seeing a group of clowns standing side by side without falling. To find the x-coordinate where the tangent line is parallel to the secant line on the interval [1, 3], we need to analyze their slopes. Once we figure out their slope relationship, finding that x-coordinate will be as easy as juggling different colored balls!
Overall, these questions have quite a mathematical twist to them. Let's see if we can untangle their mysteries together!
a. To find the zeros of f, we need to solve the equation f(x) = 0.
f(x) = x^3 - 7x + 6 = 0
To find the zeros, we can factor the equation or use numerical methods.
By observing the equation, we can see that f(1) = 0, so x = 1 is one of the zeros.
Using long division or synthetic division, we can divide f(x) by (x-1) to find the quadratic factor:
1 | 1 - 7 + 6
- 1 - 6
__________
1 - 6 + 6
- 6 + 6
__________
0
So, (x-1) is a factor of f(x), and the quotient is x^2 - 6.
Now we have (x-1)(x^2 - 6) = 0
Setting each factor equal to zero gives us:
x - 1 = 0 => x = 1
x^2 - 6 = 0
x^2 = 6
x = ±√6
Therefore, the zeros of f are x = 1, x = √6, and x = -√6.
b. To find the equation of the line tangent to the graph of f at x = -1, we need to find the derivative of f(x) and evaluate it at x = -1.
f(x) = x^3 - 7x + 6.
Taking the derivative with respect to x, we get:
f'(x) = 3x^2 - 7.
Evaluating f'(-1), we get:
f'(-1) = 3(-1)^2 - 7
= 3 - 7
= -4.
So, the slope of the tangent line at x = -1 is -4.
Now, we need to find the point on the graph of f at x = -1. Evaluating f(-1), we have:
f(-1) = (-1)^3 - 7(-1) + 6
= -1 + 7 + 6
= 12.
Therefore, the point on the graph of f at x = -1 is (-1, 12).
Using the point-slope form of the equation of a line, the equation of the tangent line can be written as:
y - y1 = m(x - x1),
where (x1, y1) is a point on the line and m is the slope.
Plugging in the values, we get:
y - 12 = -4(x - (-1))
y - 12 = -4(x + 1)
y - 12 = -4x - 4
y = -4x + 8.
Hence, the equation of the line tangent to the graph of f at x = -1 is y = -4x + 8.
c. To find the x coordinate of the point where the tangent line is parallel to the secant line on the interval [1, 3], we need to find the slope of the secant line and then find the x coordinate where the tangent line has the same slope.
The slope of the secant line passing through the points (1, f(1)) and (3, f(3)) is:
slope = (f(3) - f(1)) / (3 - 1).
Evaluating f(3) and f(1), we get:
f(3) = (3)^3 - 7(3) + 6
= 27 - 21 + 6
= 12.
f(1) = (1)^3 - 7(1) + 6
= 1 - 7 + 6
= 0.
Therefore, the slope of the secant line is:
slope = (12 - 0) / (3 - 1)
= 12 / 2
= 6.
For the tangent line to be parallel, it should have the same slope (-4) as the secant line. So, we need to find the x coordinate where the tangent line has a slope of -4.
Setting the derivative f'(x) equal to -4, we have:
-4 = 3x^2 - 7
3x^2 = 3
x^2 = 1
x = ±1.
Hence, the x coordinate of the point where the tangent line is parallel to the secant line on the interval [1, 3] is x = ±1.
a. To find the zeros of the function f(x) = x^3 - 7x + 6, we need to find the values of x for which f(x) equals zero. In other words, we need to solve the equation x^3 - 7x + 6 = 0.
Unfortunately, there is no simple formula to solve cubic equations like this one. However, we can use numerical methods or guess-and-check to find the zeros.
One way to start is by trying different values of x and see if they make the equation equal to zero. For example, we can try x = 0:
f(0) = (0)^3 - 7(0) + 6
= 0 - 0 + 6
= 6, which is not equal to zero.
We can try a few more values and see if any of them yield f(x) = 0:
x = -1: f(-1) = (-1)^3 - 7(-1) + 6 = -1 + 7 + 6 = 12, not zero.
x = 1: f(1) = (1)^3 - 7(1) + 6 = 1 - 7 + 6 = 0.
So, we have found one zero: x = 1.
Now we need to solve the quadratic equation for the remaining zeros. One possible method is to use polynomial long division to divide f(x) by x - 1:
x^2 + x - 6
________________
x - 1 | x^3 - 7x + 6
This division results in a quotient of x^2 + x - 6. Now we need to factor this quadratic equation:
(x - 2)(x + 3) = 0.
Setting each factor equal to zero, we find x = 2 and x = -3 as the remaining zeros.
Therefore, the zeros of f are x = 1, x = 2, and x = -3.
b. To find the equation of the line tangent to the graph of f at x = -1, we need to find the slope of the tangent line and a point on the line.
The slope of the tangent line can be found using the derivative of f(x). The derivative of f(x) = x^3 - 7x + 6 is given by:
f'(x) = 3x^2 - 7.
Plugging in x = -1 into the derivative, we get:
f'(-1) = 3(-1)^2 - 7
= 3 - 7
= -4.
This is the slope of the tangent line.
To find a point on the tangent line, we need to find the corresponding y-value at x = -1. Plugging x = -1 into f(x), we have:
f(-1) = (-1)^3 - 7(-1) + 6
= -1 + 7 + 6
= 12.
So, a point on the tangent line is (-1, 12).
Now, we can use the point-slope form of a line to write the equation of the tangent line:
y - y1 = m(x - x1),
where (x1, y1) is a point on the line and m is the slope.
Plugging in the values (-1, 12) and -4 for x1, y1, and m, respectively, we have:
y - 12 = -4(x - (-1))
= -4(x + 1).
Expanding this equation yields the equation of the tangent line:
y - 12 = -4x - 4.
Thus, the equation of the line tangent to the graph of f at x = -1 is y = -4x + 8.
c. To find the x-coordinate where the tangent line is parallel to the secant line on the interval [1, 3], we need to find the slope of the secant line and compare it with the slope of the tangent line.
The slope of the secant line can be found using the formula:
m_secant = (f(3) - f(1))/(3 - 1).
Plugging in the values f(3) and f(1), we have:
m_secant = (3^3 - 7(3) + 6 - (1^3 - 7(1) + 6))/(3 - 1)
= (27 - 21 + 6 - (1 - 7 + 6))/(3 - 1)
= (12 - 4)/(3 - 1)
= 8/2
= 4.
So, the slope of the secant line is 4.
Since the tangent line is parallel to the secant line, their slopes must be equal. We know from part b that the slope of the tangent line is -4.
Setting the slope of the tangent line equal to the slope of the secant line, we have:
-4 = 4.
This equation has no solution. Therefore, there is no x-coordinate where the tangent line is parallel to the secant line on the interval [1, 3].