The curve f(x)=-x^4+2x^2+x has a tangent at the point (1,2). Is this line tangent to the curve at another point and, if so where?
dy/dx = 4x^3 + 4x + 1
so at (1,2) dy/dx = 4+4+1 = 9
so is 4x^3 + 4x + 1 = 9 ?
x^3 + x - 2 = 0
we already know that (x-1) is a factor, so by synthetic division I found
x^3 + x - 2 = 0
(x-1)(x^2 + x + 2) = 0
x = 1, we knew that or
x = (-1±√-7)/2 which is not real
so there are no other points where the tangent has a slope of 9
find the tangent line, graph both of them
To determine if the line tangent to the curve at the point (1,2) is also tangent at another point, we need to find the derivative of the curve and then check if there is any other point that satisfies the equation of the tangent line.
1. Find the derivative of the curve:
The derivative of f(x) can be found by applying the power rule for differentiation. Differentiating each term individually, we get:
f'(x) = -4x^3 + 4x + 1
2. Determine the equation of the tangent line at (1,2):
Since the tangent line passes through the point (1,2), we can use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope of the tangent line.
Substituting the coordinates of (1,2), we have:
y - 2 = m(x - 1)
3. Find the slope of the tangent line at (1,2):
To find the slope, substitute x = 1 into the derivative f'(x):
f'(1) = -4(1)^3 + 4(1) + 1
= -4 + 4 + 1
= 1
Therefore, the slope of the tangent line is m = 1.
4. Rewrite the equation of the tangent line using the slope-intercept form:
y - 2 = 1(x - 1)
y - 2 = x - 1
y = x + 1
So, the equation of the tangent line at (1,2) is y = x + 1.
5. Determine if the line is tangent at another point:
To find if the line is tangent at another point, we need to solve the equation of the line y = x + 1 simultaneously with the equation of the curve f(x) = -x^4 + 2x^2 + x:
-x^4 + 2x^2 + x = x + 1
Rearranging the equation and simplifying, we obtain:
-x^4 + 2x^2 = 0
Factoring out an x^2 term, we have:
x^2(-x^2 + 2) = 0
Setting each factor equal to zero:
x^2 = 0 --> x = 0
-x^2 + 2 = 0 --> x^2 = 2 --> x = ±√(2)
Therefore, the line y = x + 1 is tangent to the curve at the points (0,1) and (√2, √2 + 1) or (-√2, -√2 + 1).
In conclusion, the line y = x + 1 is tangent to the curve f(x) = -x^4 + 2x^2 + x at the points (0,1), (√2, √2 + 1), and (-√2, -√2 + 1).