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Chapter 4: Problem 21

Let \(f:[a, b] \rightarrow \mathbb{R}\) be continuous on \([a, b]\) and differentiable on \((a, b) .\) If \(f(a)0\) for some \(c \in(a, b)\).

Short Answer

Expert verified
By applying the Mean Value Theorem to the continuous function \(f\) over the interval \([a, b]\) and differentiable on \((a, b)\), we find some \(c \in (a, b)\) such that \(f'(c) = \frac{f(b)-f(a)}{b-a}\). Since \(f(a) < f(b)\) and \(b > a\), we have \(f'(c) > 0\).

Step by step solution

01

Recall the Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function \(g\) is continuous over an interval \([p, q]\) and differentiable on \((p, q)\), then there exists some \(d \in (p, q)\) such that \[g'(d)=\frac{g(q)-g(p)}{q-p}\].
02

Set up the Mean Value Theorem with our function

We will apply the MVT to our function \(f\) over the interval \([a, b]\). We know that \(f\) is continuous over \([a, b]\) and differentiable on \((a, b)\), so by the MVT, there exists some \(c \in (a, b)\) such that: \[f'(c)=\frac{f(b)-f(a)}{b-a}\].
03

Prove \(f'(c)>0\)

Since we know that \(f(a) < f(b)\), it follows that \(f(b)-f(a) > 0\). Also, we have \(b > a\), and therefore \(b - a > 0\). Now, returning to the equation from the MVT, \[f'(c)=\frac{f(b)-f(a)}{b-a}\]. Since both the numerator and the denominator are positive \((f(b) - f(a) > 0\) and \(b - a > 0\)), it follows that their quotient is also positive, \[f'(c)>0\]. We have shown that there exists some \(c \in (a, b)\) such that \(f'(c) > 0\).

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Most popular questions from this chapter

(i) Let \(f, g:[a, b] \rightarrow \mathbb{R}\) be continuous on \([a, b]\) and differentiable on \((a, b)\). If \(f(a) \leq g(a)\) and \(f^{\prime}(x) \leq g^{\prime}(x)\) for all \(x \in(a, b)\), then show that \(f(b) \leq g(b)\) (ii) Use (i) to show that \(15 x^{2} \leq 8 x^{3}+12 \leq 18 x^{2}\) for all \(x \in[1.25,1.5]\). Deduce that the range of the function \(h:[1.25,1.5] \rightarrow \mathbb{R}\) given by \(h(x)=\left(2 x^{3}+3\right) / 3 x^{2}\) is contained in \([1.25,1.5]\).

Find the points on the curve \(x^{2}+x y+y^{2}=7\) at which (i) the tangent is parallel to the \(x\) -axis, (ii) the tangent is parallel to the \(y\) -axis.

Let \(I\) be an interval containing more than one point and \(f: I \rightarrow \mathbb{R}\) be a convex function. (i) Show that for every interior point \(c\) of \(I\), both \(f_{-}^{\prime}(c)\) and \(f_{+}^{\prime}(c)\) exist and \(f_{-}^{\prime}(c) \leq f_{+}^{\prime}(c) .\) (Hint: Use Exercise 72 of Chapter \(1 .\) ) (ii) Show that for any \(x_{1}, x_{2} \in I\) with \(x_{1}

Consider \(f: \mathbb{R} \backslash\\{1\\} \rightarrow \mathbb{R}\) and \(g: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x):=\frac{1}{x-1}\) for \(x \neq 1\) and \(g(x):=x\) for \(x \in \mathbb{R}\). Show that $$ \frac{f^{\prime}(x)}{g^{\prime}(x)} \rightarrow-\infty \text { as } x \rightarrow 1^{+}, \quad \text { but } \quad \frac{f(x)}{g(x)} \rightarrow \infty \text { as } x \rightarrow 1^{+} . $$ Does this contradict L'H么pital's Rule? Justify.

Let \(f:(a, b) \rightarrow \mathbb{R}\) and \(c \in(a, b) .\) Show that the following are equivalent: (i) \(f\) is differentiable at \(c\). (ii) There exist \(\alpha \in \mathbb{R}, \delta>0\) and a function \(\epsilon_{1}:(-\delta, \delta) \rightarrow \mathbb{R}\) such that \(f(c+h)=f(c)+\alpha h+h \epsilon_{1}(h)\) for all \(h \in(-\delta, \delta)\) and \(\lim _{h \rightarrow 0} \epsilon_{1}(h)=0\) (iii) There exists \(\alpha \in \mathbb{R}\) such that $$ \lim _{h \rightarrow 0} \frac{|f(c+h)-f(c)-\alpha h|}{|h|}=0 $$ If the above conditions hold, then show that \(f^{\prime}(c)=\alpha\).
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