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Chapter 8: Problem 18

Find the equation of the common tangent to the curves \(y=x^{2}-5 x+4\) and \(y=x^{2}+x+1\).

Short Answer

Expert verified
There is no common tangent to the given curves.

Step by step solution

01

Calculate the derivatives

Differentiate both of the quadratic functions. The derivative of the first function, \(y=x^{2}-5 x+4\), using power rule results in \(y'=2x-5\). Similarly, the derivative of the second function, \(y=x^{2}+x+1\), also using the power rule, gives \(y'=2x+1\).
02

Set the derivatives equal

The condition for a common tangent to two curves is equal slopes. Set \(y_1'=y_2'\). Hence, \(2x-5=2x+1\). Solving this we find that there is no x-value satisfying this equation.
03

Conclusion

Since there cannot be a common x-value and hence no common tangent, the solution to this problem is that there is not any common tangent to the given curves.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative Rules
When dealing with curves and attempting to find tangents, understanding derivative rules is essential. Derivatives help us find slopes of tangent lines at specific points on a curve. The power rule is one of the most straightforward and commonly used rules in calculus. This rule states that for a function of the form \( y = ax^n \), its derivative is \( y' = nax^{n-1} \). Using this rule:
  • The derivative of \( y = x^2 - 5x + 4 \) is found by differentiating each term separately. The derivative of \( x^2 \) is \( 2x \), for \( -5x \) is \( -5 \), and for the constant \( 4 \) is \( 0 \). Hence, the derivative is \( y' = 2x - 5 \).
  • Similarly, for the function \( y = x^2 + x + 1 \), the derivative is \( y' = 2x + 1 \).
Knowing these derivatives is crucial as they represent the slope of the tangent lines at any point \( x \) on their respective curves.
Tangent to a Curve
A tangent to a curve is a straight line that touches the curve at just one point. This means that at this point of contact, the slope of the tangent line is exactly the slope of the curve. To find a common tangent between two curves, we need to locate a place where both curves have the same slope. This boils down to setting their derivatives equal. However, if upon solving, the equation does not yield a solution, it implies that no such common tangent exists.
In this specific exercise, by equating the derivatives \( 2x - 5 = 2x + 1 \), it became evident that no common \( x \)-value could satisfy this equation. Therefore, these curves do not share a common tangent.
Quadratic Functions
Quadratic functions are a type of polynomial that take the form \( y = ax^2 + bx + c \). The graph of a quadratic function is a parabola, which can either open upwards or downwards depending on the sign of the coefficient \( a \). Some key properties include:
  • The vertex, which represents the maximum or minimum point on the graph.
  • The axis of symmetry which is a vertical line passing through the vertex dividing the parabola into two symmetric halves.
  • The zeros or x-intercepts, which are the points where the graph crosses the x-axis.
For the functions given in the problem, both \( y = x^2 - 5x + 4 \) and \( y = x^2 + x + 1 \) are quadratic. Each has a distinct vertex and may intersect the x-axis at different places, but in this case, an exploration of their slopes, derived from their derivatives, led to the conclusion that they do not possess a common tangent.

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Most popular questions from this chapter

The normal at any point \(P\left(c t, \frac{c}{t}\right)\) on the curve \(x y=c^{2}\) meets the curve at \(Q\left(c t_{1}, \frac{c}{t_{1}}\right)\) then \(t_{1}\) is (a) \(-t\) (b) \(\frac{1}{t^{2}}\) (c) \(-\frac{1}{t^{3}}\) (d) None

Find the angle between the curves \(y^{2}=4 x\) and \(y=e^{-x / 2}\)

Find a point on the curve \(x^{2}+2 y^{2}=6\) whose distance from the line \(x+y=7\) is minimum.

The point on the curve where the normal to the curve \(9 y^{2}=x^{3}\) makes equal intercepts with the axes is (a) \(\left(4, \frac{8}{3}\right)\) (b) \(\left(-4, \frac{8}{3}\right)\) (c) \(\left(4,-\frac{8}{3}\right)\) (d) None

If the line joining the points \((0,3)\) and \((5,-2)\) is a tangent to the curve \(y=\frac{a x}{x+1}\), then the value of \(a\) is/are (a) \(a=1 \pm \sqrt{3}\) (b) \(a=2 \pm 2 \sqrt{3}\) (c) \(a=-1 \pm 1 \sqrt{3}\) (d) \(a=2-\sqrt{3}\)
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