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Accedmychevron_right11thchevron_rightmathchevron_rightFunctions And Graphschevron_rightExercise 2.1

Solution

We simplify:

f(a+h)−f(a)h\frac{f(a+h)-f(a)}{h}

(i) f(x)=4x+7f(x)=4x+7

f(a+h)=4(a+h)+7=4a+4h+7f(a)=4a+7\begin{aligned} f(a+h)&=4(a+h)+7=4a+4h+7\\ f(a)&=4a+7 \end{aligned} f(a+h)−f(a)h=(4a+4h+7)−(4a+7)h=4hh=4\frac{f(a+h)-f(a)}{h}=\frac{(4a+4h+7)-(4a+7)}{h}=\frac{4h}{h}=4 4\boxed{4}

(ii) f(x)=sin⁡xf(x)=\sin x

f(a+h)−f(a)h=sin⁡(a+h)−sin⁡ah\frac{f(a+h)-f(a)}{h}=\frac{\sin(a+h)-\sin a}{h}

Using sin⁡u−sin⁡v=2cos⁡(u+v2)sin⁡(u−v2)\sin u-\sin v=2\cos\left(\frac{u+v}{2}\right)\sin\left(\frac{u-v}{2}\right):

sin⁡(a+h)−sin⁡a=2cos⁡(a+h2)sin⁡(h2)\sin(a+h)-\sin a=2\cos\left(a+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)

So:

sin⁡(a+h)−sin⁡ah=cos⁡(a+h2)⋅sin⁡(h/2)h/2\frac{\sin(a+h)-\sin a}{h}=\cos\left(a+\frac{h}{2}\right)\cdot\frac{\sin(h/2)}{h/2} cos⁡(a+h2)⋅sin⁡(h/2)h/2\boxed{\cos\left(a+\frac{h}{2}\right)\cdot\frac{\sin(h/2)}{h/2}}

(iii) f(x)=x2+x2−1=2x2−1f(x)=x^2+x^2-1=2x^2-1

f(a+h)=2(a+h)2−1=2(a2+2ah+h2)−1=2a2+4ah+2h2−1f(a)=2a2−1\begin{aligned} f(a+h)&=2(a+h)^2-1=2(a^2+2ah+h^2)-1\\ &=2a^2+4ah+2h^2-1\\ f(a)&=2a^2-1 \end{aligned} f(a+h)−f(a)h=(2a2+4ah+2h2−1)−(2a2−1)h=4ah+2h2h=4a+2h\frac{f(a+h)-f(a)}{h}=\frac{(2a^2+4ah+2h^2-1)-(2a^2-1)}{h}=\frac{4ah+2h^2}{h}=4a+2h 4a+2h\boxed{4a+2h}

(iv) f(x)=tan⁡xf(x)=\tan x

f(a+h)−f(a)h=tan⁡(a+h)−tan⁡ah\frac{f(a+h)-f(a)}{h}=\frac{\tan(a+h)-\tan a}{h}

Using tan⁡α−tan⁡β=sin⁡(α−β)cos⁡αcos⁡β\tan\alpha-\tan\beta=\dfrac{\sin(\alpha-\beta)}{\cos\alpha\cos\beta}:

tan⁡(a+h)−tan⁡a=sin⁡hcos⁡(a+h)cos⁡a\tan(a+h)-\tan a=\frac{\sin h}{\cos(a+h)\cos a}

So:

tan⁡(a+h)−tan⁡ah=sin⁡hhcos⁡(a+h)cos⁡a\boxed{\frac{\tan(a+h)-\tan a}{h}=\frac{\sin h}{h\cos(a+h)\cos a}}
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