Question

Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

Answer 1

Answer:

a)

$g(2.9) \approx -6.6$

$g(3.1) \approx -3.4$

b)

The values are too small since $g''$ is positive for both values of $x$ in. I'm speaking of the $x$ values, 2.9 and 3.1.

Step-by-step explanation:

a)

The point-slope of a line is:

$y-y_1=m(x-x_1)$

where $m$ is the slope and $(x_1,y_1)$ is a point on that line.

We want to find the equation of the tangent line of the curve $g$ at the point $(3,-5)$ on $g$.

So we know $(x_1,y_1)=(3,-5)$.

To find $m$, we must calculate the derivative of $g$ at $x=3$:

$m=g'(3)=(3)^2+7=9+7=16$.

So the equation of the tangent line to curve $g$ at $(3,-5)$ is:

$y-(-5)=16(x-3)$.

I'm going to solve this for $y$.

$y-(-5)=16(x-3)$

$y+5=16(x-3)$

Subtract 5 on both sides:

$y=16(x-3)-5$

What this means is for values $x$ near $x=3$ is that:

$g(x) \approx 16(x-3)-5$.

Let's evaluate this approximation function for $g(2.9)$.

$g(2.9) \approx 16(2.9-3)-5$

$g(2.9) \approx 16(-.1)-5$

$g(2.9) \approx -1.6-5$

$g(2.9) \approx -6.6$

Let's evaluate this approximation function for $g(3.1)$.

$g(3.1) \approx 16(3.1-3)-5$

$g(3.1) \approx 16(.1)-5$

$g(3.1) \approx 1.6-5$

$g(3.1) \approx -3.4$

b) To determine if these are over approximations or under approximations I will require the second derivative.

If $g''$ is positive, then it leads to underestimation (since the curve is concave up at that number).

If $g''$ is negative, then it leads to overestimation (since the curve is concave down at that number).

$g'(x)=x^2+7$

$g''(x)=2x+0$

$g''(x)=2x$

$2x$ is positive for $x>0$.

$2x$ is negative for $x$.

That is, $g''(2.9)>0 \text{ and } g''(3.1)>0$.

So $2x$ is positive for both values of $x$ which means that the values we found in part (a) are underestimations.