Problem 65
Question
Let \(f\) be \(a\) continuous, diferentiable and bijective function. If the tangent to \(y=f(x)\) at \(x=a\) is also the normal to \(y=f(x)\) at \(x=b\), then there exists at least one \(c \in(a, b)\) such that (A) \(f^{\prime}(c)=0\) (B) \(f^{\prime}(c)>0\) (C) \(f^{\prime}(c)<0\) (D) None of these
Step-by-Step Solution
Verified Answer
(A) There exists a point \(c\) such that \(f'(c) = 0\).
1Step 1: Identify Key Properties
The function \(f\) is continuous, differentiable, and bijective. This means \(f\) has an inverse, \(f^{-1}\), and both \(f\) and \(f^{-1}\) are continuous and differentiable. Bijectivity also implies \(f'(x) eq 0\) for all \(x\).
2Step 2: Analyze Tangent Line at \(x = a\)
The tangent line to \(y = f(x)\) at \(x = a\) has a slope equal to \(f'(a)\). Therefore, the equation of the tangent line at \(x = a\) is \(y - f(a) = f'(a)(x - a)\). This line is also given to be the normal to the curve at \(x = b\).
3Step 3: Analyze Normal Line at \(x = b\)
The normal line to \(y = f(x)\) at \(x = b\) is perpendicular to the tangent at that point. Therefore, the slope of this normal line is the negative reciprocal of the slope of the tangent line at \(x = b\), i.e., \(-\frac{1}{f'(b)}\). Since this normal line is given to be the tangent at \(x = a\), we set \(f'(a) = -\frac{1}{f'(b)}\).
4Step 4: Set up the Equation
From the previous step, we have the equation \(f'(a) = -\frac{1}{f'(b)}\). This implies that \(f'(a) \cdot f'(b) = -1\).
5Step 5: Use Intermediate Value Theorem (IVT)
Since \(f\) is differentiable and its derivative is continuous, by IVT, between any two points \(a\) and \(b\), \(f'(x)\) takes all values between \(f'(a)\) and \(f'(b)\). Given \(f'(a) \cdot f'(b) = -1\), the product changes its sign from positive to negative (or vice-versa), implying there exists a \(c \in (a, b)\) such that \(f'(c) = 0\).
6Step 6: Conclusion
From step 5, the conclusion is that there exists at least one point \(c\) in the interval \((a, b)\) such that \(f'(c) = 0\). Thus, the answer is (A) \(f'(c) = 0\).
Key Concepts
Intermediate Value TheoremTangent and Normal LinesContinuous and Differentiable Functions
Intermediate Value Theorem
The Intermediate Value Theorem (IVT) is a fundamental concept in calculus. It states that for any continuous function \( f \) that spans between two values \( f(a) \) and \( f(b) \) on a closed interval \([a, b]\), the function will take on every value between \( f(a) \) and \( f(b) \) at some point within that interval.
This theorem is crucial in proving the existence of a root or a specific value within an interval. It is applicable when you need to find out if a function crosses a particular value within a specific range.
In our exercise, IVT helps confirm that if \( f'(a) \cdot f'(b) = -1 \), then there's a point \( c \) where the derivative \( f'(c) = 0 \). This occurs because the IVT ensures that \( f'(x) \), a continuous function, must pass through all intermediate values between \( f'(a) \) and \( f'(b) \) as it moves from one end of the interval to the other. Thus, if the sign changes, zero is one of those values.
This theorem is crucial in proving the existence of a root or a specific value within an interval. It is applicable when you need to find out if a function crosses a particular value within a specific range.
In our exercise, IVT helps confirm that if \( f'(a) \cdot f'(b) = -1 \), then there's a point \( c \) where the derivative \( f'(c) = 0 \). This occurs because the IVT ensures that \( f'(x) \), a continuous function, must pass through all intermediate values between \( f'(a) \) and \( f'(b) \) as it moves from one end of the interval to the other. Thus, if the sign changes, zero is one of those values.
Tangent and Normal Lines
Understanding tangent and normal lines is essential in differential calculus. A tangent line to a curve at a given point has the same slope as the curve at that point. If \( y = f(x) \), the slope of the tangent line at any point \( x = a \) is the derivative \( f'(a) \).
The equation of a tangent line at \( x = a \) can is given by:
On the other hand, a normal line is perpendicular to the tangent line at a given point on the curve. This means the slope of the normal line at point \( x = b \) is the negative reciprocal of the slope of the tangent line, which is \(-\frac{1}{f'(b)}\).
In our exercise, the tangent line at \( x = a \) is also the normal line at \( x = b \). Hence we establish the unique condition :
The equation of a tangent line at \( x = a \) can is given by:
- \( y - f(a) = f'(a)(x - a) \)
On the other hand, a normal line is perpendicular to the tangent line at a given point on the curve. This means the slope of the normal line at point \( x = b \) is the negative reciprocal of the slope of the tangent line, which is \(-\frac{1}{f'(b)}\).
In our exercise, the tangent line at \( x = a \) is also the normal line at \( x = b \). Hence we establish the unique condition :
- \( f'(a) = -\frac{1}{f'(b)} \), leading to \( f'(a) \cdot f'(b) = -1 \)
Continuous and Differentiable Functions
Continuous and differentiable functions have smooth graphs without breaks or sharp turns.
A function \( f \) is continuous at a point if the limit of \( f \) as it approaches that point from both directions is the same as the function's value at that point.
Differentiability relates to continuity but adds the requirement that the function's rate of change (slope) is consistent and predictable. This is expressed using derivatives. If a function is differentiable at a point, it is also continuous at that point. However, a function can be continuous but not differentiable.
In the exercise, \( f \) being continuous and differentiable ensures the Intermediate Value Theorem's applicability. It confirms that \( f'(x) \), the derivative, is continuous. Thus, any changes in \( f'(x) \) across interval \((a, b)\) are smooth, allowing us to identify at least one point \( c \) within \((a, b)\) where \( f'(c) = 0 \). This continuity ensures \( f \) does not "jump over" potential roots or slopes.
A function \( f \) is continuous at a point if the limit of \( f \) as it approaches that point from both directions is the same as the function's value at that point.
Differentiability relates to continuity but adds the requirement that the function's rate of change (slope) is consistent and predictable. This is expressed using derivatives. If a function is differentiable at a point, it is also continuous at that point. However, a function can be continuous but not differentiable.
In the exercise, \( f \) being continuous and differentiable ensures the Intermediate Value Theorem's applicability. It confirms that \( f'(x) \), the derivative, is continuous. Thus, any changes in \( f'(x) \) across interval \((a, b)\) are smooth, allowing us to identify at least one point \( c \) within \((a, b)\) where \( f'(c) = 0 \). This continuity ensures \( f \) does not "jump over" potential roots or slopes.
Other exercises in this chapter
Problem 63
Let \(f\) be a function which is continuous and differentiable for all real \(x\). If \(f(2)=-4\) and \(f^{\prime}(x) \geq 6\) for all \(x \in[2,4]\), then (A)
View solution Problem 64
If \(a x+\frac{b}{x} \geq c\) for all positive \(x\), where \(a, b>0\), then (A) \(a b
View solution Problem 66
The values of \(k\) for which the function \(f(x)=k x^{3}-9 x^{2}+9 x+3\) may be increasing on \(R\) are (A) \(k>3\) (B) \(k
View solution Problem 67
The least possible value of \(k\) for which the function \(f(x)=x^{2}+k x+1\) may be increasing on \([1,2]\) is (A) 2 (B) \(-2\) (C) 0 (D) None of these
View solution