Math  /  Algebra

Question(a) Let g(x)g(x) be some function whose graph passes through the point (3,6)(3,-6). Let h(x)=g(x)+7h(x)=g(x)+7. Based on this information, what point do you know must be on the graph of hh ? (b) Let g(x)g(x) be some function whose domain is the interval [1,2][1,2] and whose image is the interval [1,1][-1,1]. Let h(x)=g(15x)h(x)=g\left(\frac{1}{5} x\right). What are the domain and image of h(x)h(x) ?

Studdy Solution

STEP 1

1. For part (a), g(x) g(x) passes through the point (3,6) (3, -6) .
2. h(x)=g(x)+7 h(x) = g(x) + 7 .
3. For part (b), the domain of g(x) g(x) is [1,2][1, 2] and the image is [1,1][-1, 1].
4. h(x)=g(15x) h(x) = g\left(\frac{1}{5} x\right) .

STEP 2

1. Solve part (a) by determining the point on h(x) h(x) using the transformation.
2. Solve part (b) by determining the domain of h(x) h(x) .
3. Solve part (b) by determining the image of h(x) h(x) .

STEP 3

For part (a), since h(x)=g(x)+7 h(x) = g(x) + 7 , this represents a vertical shift of the graph of g(x) g(x) upwards by 7 units.
The point (3,6) (3, -6) on g(x) g(x) becomes (3,6+7)=(3,1) (3, -6 + 7) = (3, 1) on h(x) h(x) .

STEP 4

For part (b), to find the domain of h(x)=g(15x) h(x) = g\left(\frac{1}{5} x\right) , we need to determine the values of x x such that 15x \frac{1}{5} x is within the domain of g(x) g(x) , which is [1,2][1, 2].
Solve the inequalities: 115x2 1 \leq \frac{1}{5} x \leq 2
Multiply through by 5 to clear the fraction: 5x10 5 \leq x \leq 10
Thus, the domain of h(x) h(x) is [5,10][5, 10].

STEP 5

The image of h(x)=g(15x) h(x) = g\left(\frac{1}{5} x\right) is the same as the image of g(x) g(x) because the transformation 15x \frac{1}{5} x only affects the domain, not the range.
Thus, the image of h(x) h(x) is [1,1][-1, 1].
The solutions are: (a) The point on the graph of h(x) h(x) is (3,1) (3, 1) . (b) The domain of h(x) h(x) is [5,10][5, 10] and the image is [1,1][-1, 1].

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