If f(2)=13 and f'(x) \geq 1 for 2 \leq x\leq 5, how small can f(5) possibly be?


If {eq}f(2)=13 {/eq} and {eq}f'(x) \geq 1 {/eq} for {eq}2 \leq x\leq 5 {/eq}, how small can {eq}f(5) {/eq} possibly be?

Smallest Value of a Function:

Given the derivative, we can use the concept of integration, since integration is a process of finding anti-derivatives.

The anti-derivative of a function {eq}f'(x) {/eq} is {eq}f(x) {/eq}; it can be written as: {eq}f(x) = \int f'(x) \ dx {/eq} and the integration of a constant a is {eq}\int a \ dx =ax+C {/eq}, where C is a constant.

Answer and Explanation:

We are given:

{eq}f(2) = 13 \enspace and \enspace f '(x) \geq 1 {/eq} for {eq}2 \leq x \leq 5 {/eq}.

If {eq}f'(x) = m {/eq} then integrating in terms of {eq}x {/eq}, we get {eq}f(x) = \int f'(x) \ dx = mx+c {/eq}

Given that the value of {eq}f(5) {/eq} should be the smallest possible and {eq}\enspace f '(x) \geq 1, {/eq} then

{eq}m=1. {/eq}

Then the value of the function becomes {eq}f(x) = x+c {/eq}

Next, plug in {eq}x=2 {/eq} to get the value of {eq}c: {/eq}

{eq}f(2) = 1 \cdot2+c {/eq}

{eq}\Rightarrow 13= 2+c {/eq}

{eq}\Rightarrow c = 11 {/eq}

So, {eq}f(x) {/eq} now becomes

{eq}f(x) = x + 11 {/eq}

Plug in {eq}x=5 {/eq}

{eq}f(5) = 5+11=16 {/eq}

Therefore the smallest possible value of {eq}f(5) {/eq} is {eq}16. {/eq}

Learn more about this topic:

Minimum Values: Definition & Concept

from PSAT Prep: Help and Review

Chapter 18 / Lesson 16

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