(i) Let f, g:[a, b] →ℝ be continuous on [a, b] and differentiable on (a, b). If f(a) ≤g(a) and f

(i) Let f, g:[a, b] →ℝ be continuous on [a, b] and...

(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 $[1.25,1.5]$.

(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 $[1.25,1.5]$.

Key Concepts

Mean Value Theorem

The Mean Value Theorem states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists at least one point in the open interval at which the instantaneous rate of change (the derivative) equals the average rate of change over the interval. This theorem is fundamental in transferring information about the derivative to information about the function values at the endpoints.

Comparison of Derivatives and Function Values

When two functions have derivatives that one is always less than or equal to the other, and their values at a starting point are ordered similarly, one can deduce that the inequality is maintained at the end of the interval. This concept relies on both the Mean Value Theorem and the Fundamental Theorem of Calculus, allowing us to integrate a derivative bound to yield a bound on the function itself.

Function Range and Bounding Techniques

Using inequalities between functions to establish bounds is a common technique in analysis. It involves demonstrating that if a function satisfies certain inequality relationships on an interval, then its range over that interval is contained within corresponding bounds. This is particularly useful when using auxiliary functions or transformations to restrict the possible values of a function, as seen when showing the range of a given function is contained within a specific interval.

Transcript

00:02 We have two parts, part one we have two functions f and g from a closed interval a, b to the real numbers that are continuous on the closed interval a, b and differentiable on the open interval a, b.

00:17 If f at a is less than or equal to g at a and derivative of f at x is less than or equal to derivative of g at x for all x in the open interval a, b then we will show that f at b is less than or equal to g at b.

00:36 The second part we use the first part to show that 15 x square is less than 8 x cubed plus 12 and that is less than or equal to 18 x square for all x in the closed interval from 1 .25 to 1 .5.

00:56 With that we deduce that the range of the function h defined on the closed interval 1 .25 to 1 .5 to the real numbers given by h of x equal to x cubed plus 3 over 3 x square is contained in 1 .25 1 .5.

01:13 So in other words in part two the function h map the closed interval 1 .25 1 .5 into itself or at least contain it itself.

01:28 So let's call it the range is not the real numbers but is a subset of 1 .25 1 .5.

01:38 Good so let's do part one first.

01:43 So in part one we are going to define a new function let's call that function p of x defined as f of x minus g of x.

01:59 The difference of the two functions we have here so defining p of x for x in the interval a b.

02:14 So we have these functions of course defined in the closed interval a b and it's a continuous function because it's the difference of two continuous functions and is differentiable.

02:27 So p of x is continuous on a b closed interval and differentiable on the open interval a b with the derivative of p equal to derivative of f minus derivative of g of x.

03:08 But now we know for any x in the interval a b derivative of f is less than or equal to the derivative of g.

03:16 So this is less than or equal to zero for all x in the open interval a b.

03:29 That means that this function is decreasing or at least not increasing non -increasing because the derivative is negative but can be zero at some points.

03:46 So we can say that then p of x is non -increasing on a b...

Please give Ace some feedback