# The function f is twice differentiable, and the graph of f has no points of inflection. If...

## Question:

The function {eq}f {/eq} is twice differentiable, and the graph of {eq}f {/eq} has no points of inflection. If {eq}f(6)=3, \ f'(6)=-\frac {1}{2} {/eq}, and {eq}f"(6)=-2 {/eq}, which of the following could be the value of {eq}f(7) {/eq}.

a. 2

b. 2.5

c. 2.9

d. 3

e. 4

## Increasing/Decreasing and Concavity:

1) If {eq}f'(x)>0 {/eq} on an interval {eq}I {/eq} then {eq}f(x) {/eq} is increasing on {eq}I {/eq}.

2) If {eq}f'(x)<0 {/eq} on an interval {eq}I {/eq} then {eq}f(x) {/eq} is decreasing on {eq}I. {/eq}

3) If {eq}f''(x)>0 {/eq} on an interval {eq}I {/eq}, then {eq}f(x) {/eq} is concave up on {eq}I. {/eq}

4) If {eq}f''(x)<0 {/eq} on an interval {eq}I, {/eq} then {eq}f(x) {/eq} is concave down on {eq}I. {/eq}

We also recall:

1) If {eq}f(x) {/eq} is continuous at {eq}x=a {/eq} and f changes from increasing to decreasing (decreasing to increasing) at {eq}x=a, {/eq} then {eq}f {/eq} has a local maximum (minimum) at {eq}x=a. {/eq}

If {eq}f(x) {/eq} is continuous at {eq}x=a {/eq} and {eq}f(x) {/eq} changes concavity at {eq}x=a, {/eq} then {eq}f(x) {/eq} has an inflection point at {eq}x=a. {/eq}

Let {eq}f(x) {/eq} be a twice differentiable function such that {eq}f(6)=3,\, f'(6)=-\frac{1}{2} \text{ and } f''(6)=-2 {/eq}. Since {eq}f(x) {/eq} does not have an inflection point, we conclude that it is concave down on the whole real line. This means that {eq}f''(x)<0 {/eq} for all {eq}x {/eq}. This means that {eq}f'(x) {/eq} is decreasing. So that the rate of change of {eq}f(x) {/eq} is decreasing as {eq}x {/eq} increases.

This means that the tangent lines will be above the actual curve. The line tangent to {eq}y=f(x) {/eq} at {eq}x=6 {/eq} is

{eq}y=f'(6)(x-6)+f(6)=-\frac{1}{2}(x-6)+3 {/eq}

Evaluating at {eq}x=7 {/eq} we have

{eq}y=\frac{-1}{2}(7-6)+3=2.5 {/eq}

Since the actual curve is below this we conclude that {eq}f(7)<2.5 {/eq}. Therefore, (a) is the only possible answer. 