Video: Finding the Rate of Change of a Function at a Point and then Finding the Angle the Tangent Makes with the Positive π‘₯-Axis

Find the rate of change of 𝑓(π‘₯) = 4π‘₯Β² βˆ’ π‘₯ βˆ’ 3 when π‘₯ = 1 and determine, to the nearest minute, the positive angle between the tangent at (1, 0) and the positive π‘₯-axis.

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Video Transcript

Find the rate of change of 𝑓 of π‘₯ equals four π‘₯ squared minus π‘₯ minus three when π‘₯ equals one and determine to the nearest minute the positive angle between the tangent at one, zero and the positive π‘₯-axis.

Remember, we can find the rate of change of some function 𝑓 given as 𝑓 prime at a point π‘Ž by evaluating the limit as β„Ž approaches zero of 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž all over β„Ž. In this question, we’re trying to find the rate of change of the function given by four π‘₯ squared minus π‘₯ minus three when π‘₯ is equal to one. So we know 𝑓 of π‘₯ and we’re going to let π‘Ž be equal to one. This means the rate of change is given by 𝑓 prime of one. It’s the limit as β„Ž approaches zero of 𝑓 of one plus β„Ž minus 𝑓 of one all over β„Ž.

To evaluate our limit, we’ll simply begin by working out what 𝑓 of one plus β„Ž actually is. And to do so, we’ll replace each instance of π‘₯ in our original function with one plus β„Ž. That’s four times one plus β„Ž all squared minus one plus β„Ž minus three. We distribute some of our parentheses and we get four times one plus two β„Ž plus β„Ž squared minus one minus β„Ž minus three. And then, we distribute our second set of parentheses. And we get four plus eight β„Ž plus four β„Ž squared minus β„Ž minus four. Of course, four minus four is zero. So this simplifies quite nicely to four β„Ž squared plus seven β„Ž.

Our next step is to evaluate 𝑓 of one. This time, we replace each instance of π‘₯ with one. That gives us four times one squared minus one minus three, which is simply zero. So we need to work out the limit as β„Ž approaches zero of four β„Ž squared plus seven β„Ž minus zero all over β„Ž. And of course, we actually don’t need minus zero. So it’s the limit as β„Ž approaches zero of four β„Ž squared plus seven β„Ž over β„Ž.

Next, we notice that we can divide the numerator by β„Ž. And so, the rate of change of our function when π‘₯ is equal to one is going to be found by evaluating the limit as β„Ž approaches zero of four β„Ž plus seven. At this point, we can practise direct substitution: when β„Ž is equal to zero, four β„Ž is zero. So the limit as β„Ž approaches zero of four β„Ž plus seven is simply seven. So we found the rate of change of our function.

Next, we need to work out the positive angle between the tangent at one, zero and the positive π‘₯-axis. Well, by finding the rate of change at π‘₯ equals one, we found the slope of the tangent to the curve at the point where π‘₯ equals one, at the point one, zero. If we use the fact that slope is rise over run or change in 𝑦 divided by change in π‘₯, we can construct a right-angled triangle, as shown. Here, the hypotenuse represents the tangent. And we’re looking to find the angle πœƒ. It’s the angle between that tangent and the positive π‘₯-axis.

So we use the standard convention for labelling a right-angled triangle. And we see that we know the length of the opposite and the adjacent side. Since tan of πœƒ is equal to opposite divided by adjacent, we can say here that tan πœƒ is equal to seven divided by one, which is simply seven. We solve this equation by finding the inverse tan of both sides, which is 81.869 and so on. We can use the relevant button on our calculator to convert this into degrees, minutes, and seconds. And we see correct to the nearest minute, the positive angle between the tangent at one, zero and the positive π‘₯-axis is 81 degrees and 52 minutes.

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