Examples of Curve Sketching

Calculus: Graphing Using Derivatives

Tip: This isn't the place to ask a question because the teacher can't reply.

Key Questions

• Use differentiation to identify turning points.
Input f(0) to find y intercept. ( 29 in this case )
Use factor theorem to help you locate zeroes ; that is test for values that result in f(a) = 0.
Test either side of zeroes and turning points to confirm behaviour.
A quartic function is even, and this one is positive so will open upwards.
. The shape will generally look something like :

image courtesy Wikipedia.

• Information from $f \left(x\right)$

$f \left(0\right) = \frac{1}{1 + 1} = \frac{1}{2} R i g h t a r r o w$ y-intercept: $\frac{1}{2}$

$f \left(x\right) > 0 R i g h t a r r o w$ x-intercept: none

${\lim}_{x \to \infty} {e}^{x} / \left\{1 + {e}^{x}\right\} = 1 R i g h t a r r o w$ H.A.: $y = 1$

${\lim}_{x \to - \infty} {e}^{x} / \left\{1 + {e}^{x}\right\} = 0 R i g h t a r r o w$ H.A.: $x = 0$

So far we have the y-intercept (in blue) and H.A.'s (in green):

Information from $f ' \left(x\right)$

$f ' \left(x\right) = \frac{{e}^{x} \cdot \left(1 + {e}^{x}\right) - {e}^{x} \cdot {e}^{x}}{{\left(1 + {e}^{x}\right)}^{2}} = {e}^{x} / {\left(1 + {e}^{x}\right)}^{2} > 0$

$R i g h t a r r o w$ $f$ is always increasing.

Information from $f ' ' \left(x\right)$

$f ' ' \left(x\right) = \frac{{e}^{x} \cdot {\left(1 + {e}^{x}\right)}^{2} - {e}^{x} \cdot 2 \left(1 + {e}^{x}\right) {e}^{x}}{{\left(1 + {e}^{x}\right)}^{4}}$

$= \frac{{e}^{x} \left(1 + {e}^{x}\right) \left(1 - {e}^{x}\right)}{{\left(1 + {e}^{x}\right)}^{4}} = \frac{{e}^{x} \left(1 - {e}^{x}\right)}{{\left(1 + {e}^{x}\right)}^{3}}$

$f ' ' \left(x\right) > 0$ on $\left(- \infty , 0\right)$ and $f ' ' \left(x\right) < 0$ on $\left(0 , \infty\right)$

$f$ is concave upward on $\left(- \infty , 0\right)$ and downward on $\left(0 , \infty\right)$.

Hence, we have the graph of $f$ (in blue):

Questions

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