Video Transcript
Use derivatives to identify which
of the following is the graph of the function 𝑓 of 𝑥 is equal to one over 10𝑥
squared plus 10𝑥.
In order to identify the graph of a
function, we usually begin by trying to find the 𝑥- and 𝑦-intercepts. However, this rational function 𝑓
of 𝑥 will not be equal to zero or have a solution at 𝑥 equals zero, and we can
therefore conclude there will be no 𝑥- or 𝑦-intercepts. This ties in with the five options
we are given, as none of the graphs cross the 𝑥- or 𝑦-axis.
Instead, this question asks us to
use derivatives to identify the graph of the function. We know that the derivative of 𝑓
of 𝑥, written 𝑓 prime of 𝑥, is equal to zero at any turning or stationary
points. We will therefore find an
expression for 𝑓 prime of 𝑥 and set it equal to zero.
Recalling that the expression one
over 𝑥 can be rewritten as 𝑥 to the power of negative one, we can rewrite our
function 𝑓 of 𝑥 as 10𝑥 squared plus 10𝑥 all raised to the power of negative
one. And we are now in a position to
differentiate this function. We can do this using the general
power rule, which is an extension of the chain rule. In this case, we multiply the
entire expression by the exponent and then reduce this exponent by one. This gives us negative one
multiplied by 10𝑥 squared plus 10𝑥 to the power of negative two. We then multiply that expression by
the derivative of the inner function.
Differentiating 10𝑥 squared plus
10𝑥 gives us 20𝑥 plus 10. And we can therefore conclude that
𝑓 prime of 𝑥 is equal to negative one multiplied by 10𝑥 squared plus 10𝑥 to the
power of negative two multiplied by 20𝑥 plus 10. This can be rewritten as negative
20𝑥 plus 10 over 10𝑥 squared plus 10𝑥 all squared. We can now set this expression
equal to zero.
For any fraction to be equal to
zero, we know the numerator must equal zero. This means that negative 20𝑥 plus
10 equals zero. Dividing through by negative one,
we have 20𝑥 plus 10 equals zero. We can then subtract 10 from both
sides and divide through by 20 such that 𝑥 is equal to negative one-half. The graph of our function must have
a turning or stationary point when 𝑥 is equal to negative one-half.
This is true of graph (A) and graph
(E). Graph (D) has no stationary point,
so we can eliminate this option. The turning point in graphs (B) and
(C) appears to occur around negative three-quarters and not negative one-half. We can therefore also eliminate
these options. In option (A), the turning point is
a local maximum, and the 𝑦-coordinate of this point lies below the 𝑥-axis so is
therefore negative, whereas in option (E), we have a local minimum and the
𝑦-coordinate lies above the 𝑥-axis and is therefore positive.
In order to find the 𝑦-coordinate
at the turning point, we can substitute 𝑥 equals negative one-half into our
original function. This gives us 𝑓 of 𝑥 is equal to
one over 10 multiplied by negative a half squared plus 10 multiplied by negative
one-half. This simplifies to one over five
over two minus five, which is equal to one over negative five over two and, in turn,
negative two-fifths. This means that the function 𝑓 of
𝑥 has a turning point at negative one-half, negative two-fifths.
As our 𝑦-coordinate is negative,
we can rule out option (E). And we can therefore conclude that
the graph of the function 𝑓 of 𝑥 is equal to one over 10𝑥 squared plus 10𝑥 is
graph (A).
An alternative to finding the
𝑦-coordinate would have been to have found the second derivative, as the value of
𝑓 double prime of 𝑥 tells us whether we have a local minimum or local maximum. If the second derivative is greater
than zero, our turning point is a minimum, whereas if it is less than zero, we have
a maximum. Whilst this is a common method, due
to the complexity of the function in this question, this would be more
difficult. And as a result, it made more sense
to find the coordinates of the turning points first. This enabled us to eliminate
options (B), (C), (D), and (E).
